∫(8) cosec 2 x dx = − cot x + c d (− cot x) = cosec 2 x
dx
∫(9) sec x tan x dx = sec x + c d (sec x) = sec x tan x
dx
∫(10) cosec x cot x dx = −cosec x + c d (−cosec x) = cosec x cot x
dx
∫(11) tan x dx = − log |cos x | +c = log | sec x | +c d (log cos x) = − tan x
dx
∫(12) cot x dx = log | sin x | +c = − log |cosec x | +c d (log sin x) = cot x
dx
∫(13) sec x dx = log | sec x + tan x| +c = log tan π + x + c d log(sec x + tan x) = sec x
4 2 dx
∫(14) cosec x dx = log | cosec x − cot x | +c = log tan x + c d (log | cosec x − cot x |) = cosec x
2 dx
∫(15) dx = sin −1 x + c = − cos −1 x + c d (sin −1 x) = 1 , d (cos −1 x) = −1
1− x2 dx 1− x2 dx 1− x2
∫(16) dx = sin −1 x + c = − cos −1 x +c d sin −1 x = 1 , d (cos −1 x ) = −1
a2 − x2 a a dx a a2 − x2 dx a a2 − x2
∫(17) dx = tan −1 x + c = − cot −1 x + c d (tan −1 x) = 1 1 , d (cot −1 x) = 1 −1 2
1+ x2 dx + x2 dx +x
∫(18) dx = 1 tan −1 x +c = −1 cot −1 x +c d tan −1 x = a d cot −1 x = −a
a2 + x2 a a a a dx a + dx a a2 + x2
a2 x2
∫(19) dx = sec −1 x + c = − cosec −1 x + c d (sec −1 x) = 1 d (cosec −1 x) = −1
x x2 −1 dx x2 −1 dx x2 −1
x x
∫(20) dx = 1 sec −1 x + c = −1 cosec −1 x + c d sec −1 x = a d (cosec −1 x ) = −a
x2 − a2 a a a a dx a x2 − a2 dx a x2 − a2
x x x
Note : In any of the fundamental integration formulae, if x is replaced by ax + b, then the same formulae
is applicable but we must divide by coefficient of x or derivative of (ax + b) i.e., a. In general, if
∫ ∫f(x)dx = φ(x) + c , then f (ax + b) dx = 1 φ (ax + b) + c
a
∫ sin(ax + b)dx = −1 cos(ax + b) + c, ∫ sec(ax + b)dx = 1 log | sec(ax + b) + tan(ax + b)| +c etc.
a a
Some more results:
∫(i) 1 = −1 coth −1 x +c = 1 log x−a +c, when x > a
x2 − a2 a a 2a x+a
∫(ii) a2 1 dx = 1 tanh −1 x +c = 1 log a + x +c, when x < a
− x2 a a 2a a − x
∫(iii) dx = log{| x + x2 − a2 |} + c = cos h−1 x + c
x2 − a2 a
∫(iv) dx = log{| x + x2 + a2 |} + c = sinh−1 x + c
x2 + a2 a
∫(v) a2 − x 2 dx = 1 x a2 − x2 + 1 a 2 sin −1 x + c
2 2 a
∫(vi) x2 − a 2 dx = 1 x x2 − a2 − 1 a 2 log{x + x2 − a2 } + c = 1 x x2 − a2 − 1 a 2 cosh −1 x + c
2 2 2 2 a
∫(vii) x2 + a 2 dx = 1 x x2 + a2 + 1 a 2 log{x + x2 + a2 }+ c = 1 x x2 + a2 + 1 a 2 sinh −1 x + c
2 2 2 2 a
Important Tips
• The signum function has an antiderivative on any interval which doesn't contain the point x = 0, and does not possess an anti-derivative
on any interval which contains the point.
• The antiderivative of every odd function is an even function and vice-versa
Example: 1 ∫ (x + 1) 2 dx is equal to [MP PET 2003]
x(x 2 +
Solution: (b) 1)
Example: 2
(a) log e x (b) log e x + 2 tan−1 x (c) log e 1 (d) log e {x(x 2 + 1)}
Solution: (c) x2 +1
Example: 3
∫ ∫ ∫ ∫ ∫ ∫(x + 1)2 x2 +1
Solution: (b) x(x 2 + 1) x((x 2
Example: 4 dx = + 2x dx = x2 +1 dx + 2 x(x x + 1) dx = dx + 2 dx = log e x + 2 tan−1 x
+ 1) x(x 2 + 1) x x2 +1
2
∫ ax 3 + bx 2 + c dx equals to [Rajasthan PET 2001]
x4
(c) None of these
(a) a log x + b + c +k (b) a log x + b − c +k (c) a log x − b − c +k
x2 3x3 x 3x3 x 3x3
ax 3 + bx2 + c a b c b c
x4 x x2 x4 x 3x3
∫ ∫I = dx = + + dx = a log x − − +k
∫ 1 + x + x + x 2 dx = [EAMCET 2003]
x + 1+ x
(d) 2(1 + x)3 / 2 + c
(a) 1 1+ x +c (b) 2 (1 + x)3 / 2 + c (c) 1 + x + c
2 3 [Rajasthan PET 2003]
[ ]1 + x + x + x 2 dx = 1+ x
∫ ∫ ∫x + 1 + x (x
1+ x + x dx = 1+ x dx = 2 (1 + x)3 / 2 + c
+ 1+ x) 3
∫ ( )sin4 x − cos 4 x dx =
(a) − cos 2x + c (b) − sin 2x + c (c) sin 2x + c (d) cos 2x + c
2 2 2 2
Solution: (b) ∫ ∫(sin 4 x − cos 4 x)dx = (sin 2 x − cos 2 x)(sin 2 x + cos 2 x)dx
Example: 5
Solution: (a) (sin 2 x − cos 2 x)dx = − (cos 2 x − sin 2 x)dx = − cos 2x dx − sin 2x +c
∫ ∫ ∫= = 2
∫1 + sin x dx = [Karnataka CET 2003]
4
(a) 8 sin x − cos x + c (b) sin x + cos x + c (c) 1 sin x − cos x + c (d) 8 cos x − sin x + c
8 8 8 8 8 8 8 8 8
sin x dx sin 2 x cos 2 x + 2 sin x . cos x dx sin x cos x 2
4 8 8 8 8 8 8
∫ ∫ ∫1+ = + = +
∫= sin x + cos x dx = − cos x / 8 + sin x / 8
8 8 1/8 1/8
= 8 sin x − cos x + c
8 8
Integration by Substitution
∫(1) When integrand is a function i.e., f [φ(x)]φ '(x) dx :
∫Here, we put φ(x) = t, so that φ'(x)dx = dt and in that case the integrand is reduced to f(t)dt . In this method,
the integrand is broken into two factors so that one factor can be expressed in terms of the function whose
differential coefficient is the second factor.
Example: 6 ∫ dx = [Rajasthan PET 2002]
Solution: (b) 2 x (1 + x)
(d) None of these
(a) 1 tan−1( x)+c (b) tan−1( x ) + c (c) 2 tan−1( x ) + c
2
∫I = dx , put x = t ⇒ 1 dx = dt
2 x (1 + x) 2x
∫∴ I = dt = tan−1 t + c = tan−1 x +c
1+ t2
(2) When integrand is the product of two factors such that one is the derivative of the others i.e.,
∫I = f'(x). f (x). dx
In this case we put f(x) = t and convert it into a standard integral.
Example: 7 ∫For any natural number m (x3m + x2m + xm)(2x2m + 3xm + 6)1 / mdx, (x > 0) = [IIT 2002]
m+1 m
(a) (2x 3m + 3x 2m + 6x m ) m (b) (2x 3m + 3x 2m + 6x m ) m+1
6(m + 1) 6(m + 1)
m+1 m
(c) (2x 3m + 3x 2m + 6x m ) m (d) − (2x 3m + 3x 2m + 6x m ) m+1
6(m + 1) 6(m + 1)
−
Solution: (a) ∫I = (x 3m−1 + x 2m−1 + x m−1)(2x 3m + 3x 2m + 6x m )1/ m dx
Example: 8 Put 2x 3m + 3x 2m + 6x m = t ⇒ 6m(x 3m−1 + x 2m−1 + x m−1 )dx = dt
Solution: (b)
Example: 9 1 1 t 1 +1 1
Solution: (a) 6m 6m m 6(m +
∫I = t1 / mdt +c t (m + 1) / m +c
= =
1 +1 1)
m
Again put the value of t, then we get option (a).
∫ x 2 tan −1 x 3 dx is equal to [MP PET 1999; UPSEAT 1999]
1+ x6
(a) tan −1(x 3 ) + c (b) 1 (tan −1 x 3)2 + c (c) − 1 (tan −1 x 3)2 + c (d) 1 (tan −1 x 3)2 + c
6 2 2
Put tan −1 x 3 = t
∫1 x2 dt 1 1 t2 (tan −1 x 3 )2
1+ x6 3 3 3 2 6
1+ x6
. 3x 2dx = dt ⇒ dx = ⇒ tdt = . = +c
∫ x −3 51/ x2 dx = k.51/ x2 + c, then k is
(a) − 1 (b) −2 log 5 (c) 2 (d) −2
2 log 5 log 5 log 5
Put x−2 = t ⇒ − 2x−3 dx = dt ⇒ x − 3dx = − dt
2
∫ ∫x−351/ x2 dx = − 1 5t dt = − 1 5t + c = −1 5 .51/ x2 +c . On comparing, k = 1
2 2 log e 2 log e − 2 log e
5 5
(3) Integral of a function of the form f (ax + b) : Here we put ax + b = t and convert it into standard
f(x)dx = φ(x), then, 1 b) +c
∫ ∫integral. Obviously if f (ax + b)dx = a φ (ax +
Example: 10 ∫ tan(3x − 5) sec(3x − 5)dx = [MP PET 1988]
Solution: (b)
(a) sec(3x − 5) + c (b) 1 sec(3x − 5) + c (c) tan(3x − 5) + c (d) None of these
3
∫ sec x tan x dx = sec x + c
∫∴ sec(3x − 5) tan(3x − 5)dx = sec(3x − 5) + c
3
(4) If integral of a function of the form f'(x) dx = log[ f (x)] + c
f (x)
Example: 11 ∫ a2 sin 2x dx is equal to [MNR 1987; Rajasthan PET 1991]
sin 2 x + b 2 cos 2 x
(a) b2 1 a2 log(a2 sin 2 x + b2 cos 2 x) + c (b) a2 1 b2 log(a 2 sin 2 x + b2 cos 2 x) + c
− −
(c) log(a2 sin2 x − b2 cos2 x) + c (d) None of these
Solution: (b) Put a 2 sin 2 x + b 2 cos 2 x = t ⇒ (a2 . 2 sin x cos x − b2. 2cos x sin x)dx = dt ⇒ sin 2x(a 2 − b 2 )dx = dt
sin 2x dx 1 dt 1 log t c 1 log(a 2 sin 2 x b2 cos 2 x) c
sin2 x + b2 cos2 − t − −
∫ ∫a2 = (a 2 = + = + +
x b2) a2 b2 a2 b2
Example: 12 ∫ dx [Rajasthan PET 1990, 94; MP PET 1991; Roorkee 1977; SCRA 1999]
1+ex =
(a) log(1 + e x ) (b) − log(1 + e −x ) (c) − log(1 − e −x ) (d) log(e −x + e −2x )
Solution: (b) dx e−x , Put e−x − e −x dx = dt ⇒ I = − dt = − log(1 + e −x )
Example: 13 1+ex = −x + t
∫ ∫ ∫I =
e 1 dx +1= t ⇒ = − log t
∫The value of the integral 1 + x x dx is equal to
x tan
(a) log | x cos x + sin x | +c (b) log |cos x + x |+ c (c) log |cos x + x sin x | +c (d) None of these
Solution: (c) x dx x cos x dx
x tan cos x + x sin
∫ ∫I = = x
1+ x
Put (cos x + x sin x) = t ⇒ (− sin x + x cos x + sin x)dx = dt ⇒ x cos x dx = dt
∫I = dt = log | t | +c = log |cos x+ x sin x | +c
t
∫(5) If integral of a function of the form, [ f (x)]n f' ( x )dx = [ f (x ]) n+1 +c [n ≠ −1]
n+1
Example: 14 ∫ x 1 + x 2 dx = [MP PET 1989]
Solution: (d) (a) 1 + 2x 2 + c (b) 1 + x 2 + c (c) 3(1 + x 2 )3 / 2 + c (d) 1 (1 + x 2 )3 / 2 + c
Example: 15 1+ x2 3
Solution: (d)
∫ ∫Put 1 + x 2 = t ⇒ 2xdx = dt , 1 1 t3/ 2 1
x 1+ x 2 dx = 2 t 1 / 2dt = 2 . 3/ +c = 3 (1 + x 2 )3 / 2 + c
2
∫ 2 + sin 3x .cos 3x dx = [IIT 1976]
(a) 2 2 + sin 3x + c (b) 2 (2 + sin 3x)2 / 3 + c (c) 2 (2 + sin 3x)3 / 2 + c (d) 2 (2 + sin 3x)3 / 2 + c
9 3 3 9
Put (2 + sin 3x) = t ⇒ 3 cos 3x dx = dt
∫ ∫2 + sin 3x .cos 3x dx = 1 t dt = 1 . t3 / 2 + c = 2 .(2 + sin 3x)3 / 2 + c .
3 3 3/ 9
2
∫(6) If the integral of a function of the form, f'(x) dx = 2 f(x) + c
f (x)
Example: 16 ∫ sin tan x x dx [MP PET 2002; Rajasthan PET 1985, 88; Bihar CEE 1974]
Solution: (b) x . cos
Example: 17
(a) 2 sec x + c (b) 2 tan x + c (c) 2 +c (d) 2 + c
Solution: (c) tan x sec x
tan x dx tan x dx = sin x . sec x dx = sec 2 x dx
x . cos
∫ ∫ ∫ ∫sin x =
tan x . sin x cos x tan x sin x cos x tan x
∫Put t = tan x ⇒ dt = sec 2 x dx, then it reduces to, 1 dt = 2t1 / 2 + c = 2 tan x + c
t
∫ x 5 dx [IIT 1975, 85]
1+ x3
(a) 2 (1 + x 3 )3 / 2 + c (b) 2 (1 + x 3 )3 / 2 + 2 (1 + x3)1 / 2 + c
9 9 3
(c) 2 (1 + x 3 )3 / 2 − 2 (1 + x3)1 / 2 + c (d) None of these
9 3
∫ ∫x 5 dx = x3.x2 dx . Put 1 + x 3 = t ⇒ 3x 2dx = dt ⇒ x 2dx = dt
1+ x3 1+ x3 3
1 (t − 1) dt = 1 t− 1 dt 1 t3/ 2 2 t c 2 (1 + x3)3 / 2 2 (1 + x3)1 / 2 c
3 t 3 t 3 9 3
∫ ∫= = 3 / 2 − + = − +
(7) Standard substitutions
Integrand form Substitution
x = a sinθ , x = a cosθ
(i) a2 − x2, 1 , a2 − x2 x = a tanθ or x = a sinhθ
x = a sec θ or x = a coshθ
a2 − x2 x = a tan2 θ
x = a sin2 θ
(ii) x2 + a2 , 1 , x2 + a2 x = a sec2 θ
x = a cos 2θ
x2 + a2 x = α cos2 θ + β sin2 θ
(iii) x2 − a2, 1 , x2 − a2
x2 − a2
(iv) x , a + x , x(a + x), 1
+ x x(a + x)
a x
(v) x , a − x , x(a − x) , 1
− x x(a − x)
a x
(vi) x , x − a , x(x − a), 1
− x x(x − a)
x a
(vii) a − x , a+x
a + x a−x
(viii) x −α , (x − α)(β − x), (β > α)
β −x
Example: 18 ∫ dx is equal to [Rajasthan PET 2000]
Solution: (b) (a 2 + x 2 )3 / 2
(d) None of these
Example: 19 (a) x +c (b) a 2 (a 2 x + c (c) 1 +c
Solution: (d) (a 2 + x 2 )1/ 2 + x 2 )1/ 2 a 2 (a 2 + x 2 )1/ 2
∫I = dx
(a 2 + x 2 )3 / 2
Put x = a tanθ ⇒ dx = a sec2 θ dθ
a sec2 θ dθ a sec 2 θ dθ 1 sec 2 θ dθ
(a2 + a2 tan2 θ )3 / 2 a 3 (sec 2 θ )3 / 2 a2 sec 3 θ
∫ ∫ ∫∴ I = = =
∫ ∫⇒ I = 1 dθ = 1 cosθ dθ ⇒ I = 1 sin θ +c ⇒ x +c
a2 sec θ a2 a2 a2(x 2 + a2)1 / 2
∫1 − x dx = [IIT 1971]
1 + x
(d) sin−1 x + 1 − x 2 + C
(a) sin −1 x − 1 1− x2 + C (b) sin −1 x + 1 1 − x 2 + C (c) sin−1 x − 1− x2 + C
2 2
Put x = cos 2θ , then θ = 1 cos−1 x ⇒ dx = −2 sin 2θ dθ
2
∫ ∫∴ I = −2 1 − cos 2θ . sin 2θ dθ 2 sin 2 θ / 2 cos 2 θ . sin 2θ .dθ
1 + cos 2θ ⇒ I = −2
⇒ I = −2.2 sin 2 θ dθ I = −2 (1 − cos 2θ )dθ I −2θ sin 2θ C1
∫ ∫ ∫⇒ I = −2 2
tanθ .2 sinθ cosθ dθ ⇒ ⇒ = − +
⇒ I = −2θ + sin 2θ + C1 ⇒ I = − cos −1 x + 1− x2 + C1 ⇒ I = − π + sin −1 x + 1 − x 2 + C1
2
⇒ I = sin −1 x + 1 − x 2 + C [where C = C1 − π ]
Trick : Rationalization of denominator and put 1 − x 2 = t 2 . 2
Integration by Parts .
(1) When integrand involves more than one type of functions : We may solve such integrals by a rule
which is known as integration by parts. We know that,
d dv du udv vdu
dx dx
∫ ∫ ∫dx
(uv) = u +v ⇒ d(uv) = + ⇒ d (uv) = udv + vdu
{ddxu . vdx}dx i.e., the integral of the product of two
∫ ∫ ∫ ∫If u and v are two functions of x, then
u v dx = u v dx −
I II
functions = (First function) × (Integral of second function) – Integral of {(Differentiation of first function) × (Integral of
second function)}
Integration with the help of above rule is called integration by parts. Before applying this rule proper choice of first
and second function is necessary. Normally we use the following methods :
(i) In the product of two functions, one of the function is not directly integrable (i.e.,
log| x|, sin−1 x, cos−1 x, tan−1 x ......etc), then we take it as the first function and the remaining function is taken as the
second function.
(ii) If there is no other function, then unity is taken as the second function e.g. In the integration of
∫ ∫sin−1 x dx, log x dx,1 is taken as the second function.
(iii) If both of the function are directly integrable then the first function is chosen in such a way that the derivative
of the function thus obtained under integral sign is easily integrable.
Usually, we use the following preference order for the first function. (Inverse, Logarithmic, Algebraic,
Trigonometric, exponential). This rule is simply called as “I LATE”.
Important Tips
If In = x n.e ax dx, then In x ne ax n I n−1
∫• = a − a
∫• If In = (log x)dx, then In = x log x − x
If In = 1 x dx, then In log(log x) log x (log x)2 (log x)3 + .......................
∫• log = + + 2.(2 ! ) + 3(3 ! )
∫• If In = (log x)n dx ; then In = x(log x)n − n.In−1
• Successive integration by parts can be performed when one of the functions is x n (n is positive integer) which will be successively
differentiated and the other is either of the following sin ax,cos ax,e ax ,e −ax ,(x + a)m which will be successively integrated.
∫ ∫• Chain rule : u .v dx = u v1 − u' v2 + u" v3 − u′′′v4 + ................. + (−1)n−1 un−1vn + (−1)n un .vn dx Where un stands for nth differential
coefficient of u and vn stands for nth integral of v.
Example: 20 ∫If In = (log x)n dx, then In + nIn−1 = [Karnataka CET 2003]
Solution: (a)
(a) x(log x)n (b) (x log x)n (c) (log x)n−1 (d) n(log x)n
Example: 21
Solution: (a) ∫ ∫In = (log x)n dx , ∴ In−1 = (log x)n−1dx
Example: 22
(log x)n.1 dx = (log x)n . x − n x)n−1 1 = (log x)n . x − n (log x)n−1 dx
∫ ∫ ∫Now In = (log . x . x dx
In = x(log x)n − n In−1 ∴ In + n In−1 = x(log x)n
∫If x sin x dx = −x cos x + A, then A = [MP PET 1992, 2000; Rajasthan PET 1997]
(a) sin x + constant (b) cos x + constant (c) Constant (d) None of these
∫Since x sin x dx = −x cos x + A ⇒ − x cos x + sin x + constant = −x cos x + A
Equating it, we get A = sin x + constant
∫ x sec 2 x dx = [Rajasthan PET 1996, 2003; MP PET 1987, 97]
(a) tan x + log cos x + c (b) x2 sec 2 x + log cos x + c (c) x tan x + log sec x + c (d) x tan x + log cos x + c
2
Solution: (d) ∫ ∫x sec 2 x dx = x tan x − tan x dx = x tan x + log(cos x) + c
Example: 23
Solution: (c) ∫[ f(x)g"(x) − f "(x) g(x)]dx is equal to [MP PET 2001]
Example: 24
Solution: (a) (a) f(x) (b) f '(x) g(x) − f(x) g'(x) (c) f(x) g'(x) − f '(x) g(x) (d) f(x) g'(x) + f '(x) g(x)
g'(x)
Example: 25
Solution: (c) ∫ ∫ ∫[ f(x) g"(x) − f "(x) g(x)]dx = f(x) g"(x)dx − f "(x) g(x)dx
∫ ∫= [ f(x) g'(x) − f '(x) g'(x)dx] − [g(x) f'(x) − g'(x)f'(x)dx] = f(x) g'(x) − f '(x) g(x)
∫ sin−1 x − cos−1 x dx is equal to [IIT 1983; WBJEE 1992]
sin −1 x + cos −1 x
(b) 2 [(2x − 1) sin−1 x − x(1 − x)] + x + c
(a) 2 [(2x − 1) sin−1 x + x(1 − x)] − x + c π
π
(d) None of these
[ ](c)π
2 (2x − 1) sin−1 x + x(1 − x) − x + c
∫ ∫ ∫sin−1
sin−1
x − cos−1 x dx = 2 sin−1 x dx − cos−1 x dx sin−1 x + cos−1 x = π
x + cos−1 x π 2
∫Now we solve first and second expressions separately. For first expression, sin−1 x dx
Put x = sin 2 θ ⇒ cos 2θ = 1 − 2x ⇒ dx = sin 2θ dθ
dθ −θ cos 2θ 1 cos 2θ dθ −θ cos 2θ sin 2θ + c1 (2x − 1) sin −1 x 1 x(1 − x) + c1
2 2 2 4 2 2
∫ ∫θ .sin 2θ = + = + = +
∫For second expression, cos−1 x dx
Put x = cos 2 θ ⇒ dx = −2 cosθ sinθ dθ = − sin 2θ dθ
∫ ∫cos−1 θ cos 2θ sin 2θ
xdx = − θ .sin 2θ dθ = 2 − 4 + c2
Therefore, I = 2 (2x − 1) {sin −1 x − cos−1 x}+ x(1 − x) + c
π 2
] [ ]I = 2 (2x − 1) sin−1 (2x − 1)
π 2
x − . π + x(1 − x) + c ⇒ I = 2 (2x − 1)sin−1 x+ x(1 − x) − x + c
2 π
∫ x 2dx [MNR 1989; Rajasthan PET 2000]
(x sin x + cos x)2 =
(a) sin x + cos x (b) x sin x − cos x (c) sin x − x cos x (d) None of these
x sin x + cos x x sin x + cos x x sin x + cos x
x 2dx x cos x x
(x sin x + cos x)2 = sin x + cos cos
∫ ∫Differentiation of x sin x + cos x is x cos x. Then, I = (x x)2 . x dx
∫Integrate by parts, 1 dt = − 1
t2 t
I −1 . x 1 . cos x .1 − x(− sin x) dx I −1 . x sec 2 x dx
∫ ∫∴ = x+ cos + x+ cos 2x ⇒ = x+ cos +
(x sin cos x) x (x sin cos x) x sin cos x x
I = −1 . x + sin x ⇒ I = − x + x sin 2 x + sin x cos x ⇒ I = sin x cos x − x(1 − sin 2 x) = sin x − x cos x
x+ cos cos x (x sin x + cos x)cos x (x sin x + cos x)cos x x sin x + cos x
x sin cos x x
∫ ∫(2) Integral is of the form ex{f (x) + f '(x)}dx : If the integral is of the form e x {f(x) + f '(x)}dx, then by
breaking this integral into two integrals integrate one integral by parts and keeping other integral as it is, by doing
so, we get
∫(i) e x [f(x) + f '(x)]dx = e x f(x) + c
∫(ii) emx [mf(x) + f '(x)]dx = emx f(x) + c
∫(iii) e mx f(x) + f '(x) dx = e mx f(x) + c
m m
Example: 26 ∫ e x (1 − cot x + cot 2 x)dx (b) e x cosec x + c (c) − e x cot x + c [MP PET 2002]
Solution: (c)
Example: 27 (a) e x cot x + c (d) e x cosec x + c
Solution: (b)
∫ ∫e x (cosec 2 x − cot x)dx = e x [− cot x + cosec 2 x] = −e x cot x + c
Example: 28
Solution: (b) ∫e x 1 − sin x dx [Rajasthan PET 2000]
1 − cos x
Example: 29
(a) − ex tan x + c (b) − ex cot x + c (c) − 1 ex tan x + c (d) 1 ex cot x + c
2 2 2 2 2 2
1 − sin x 1 2 sin x cos x 1 2 x x
1 − cos x 2 sin 2 2 2 2 2 2
= x − x = cosec − cot
2 2
2 sin 2
∴ ∫ ex 1− sin x dx = ∫ e x 1 cosec 2 x − cot x dx = −e x cot x + c
1− cos x 2 2 2 2
∫ e 2 tan−1 x (1 + x)2 dx = [WB JEE 1998]
1+ x2
(d) None of these
(a) x e tan−1 x + c (b) x e 2 tan−1 x + c (c) 2x e tan−1 x + c
Put tan−1 x = t ⇒ 1 dx = dt
1+ x2
e2 tan−1 x (1 + x)2 dx e2t(1 + tan t)2dt = e 2t (sec 2 t + 2 tan t)dt
1+ x2
∫ ∫ ∫∴ I = =
e 2t sec 2 t 2tan t e 2t sec 2 t. e 2t e 2t sec 2 t dt + e 2t tan t − e 2t sec 2 t dt + c
2 2
∫ ∫ ∫ ∫I = dt + − dt ⇒ I=
I = e 2t tan t + c ⇒ I = x e 2 tan−1 x + c
∫ (x + 3)e x dx is equal to [Karnataka CET 2000]
(x + 4)2
(a) ex + c (b) ex 3 + c (c) 1 +c (d) ex +c
x+4 x+ (x + 4)2 (x + 4)2
(x + 4 − 1)e x ex ex
(x + 4)2 (x + + 4)2
(x + 3)e x
∫ ∫ ∫ ∫ ∫ ∫(x + 4)2
Solution: (a) dx = dx = (x + 4)e x dx − (x ex dx = 4) dx − (x dx
(x + 4)2 + 4)2
∫ ∫ e x (x ex dx − (x ex dx + c = ex +c
+ 4)2 + 4)2 x+4
= (x + 4) +
∫(3) Integral is of the form [x f ′(x) + f(x)]dx : If the integral is of the form [x f '(x) + f(x)]dx then by
breaking this integral into two integrals, integrate one integral by parts and keeping other integral as it is, by doing
∫so, we get, [x f '(x) + f(x)]dx = x f(x) + c
Example: 30 If d f(x) = x cos x + sin x and f(0) = 2, then f(x) = [MP PET 1989]
Solution: (c) dx
Example: 31
(a) x sin x (b) x cos x + sin x + 2 (c) x sin x + 2 (d) x cos x + 2
Solution: (b)
∫d
dx
f(x) = x cos x + sin x ⇒ f(x) = (x cos x + sin x)dx = x sin x + c
Since f(0) = 2 ⇒ c = 2 , ∴ f(x) = x sin x + 2
∫ x + sin x dx = [Roorkee 1980; AMU 1997; UPSEAT 1999]
1 + cos x
(a) −x tan x / 2 + c (b) x tan x / 2 + c (c) x tan x + c (d) 1 x tan x + c
2
1 x tan x / 2 x x x
2 1/ 2 2 2 2
∫ ∫ ∫ ∫ ∫x + sin x 1 x x
1 + cos x 2 2 2
dx = x sec 2 dx + tan dx = − tan dx + tan dx = x tan + c
Trick : By inspection, d x tan x + c = x sec 2 x + tan x = 1 x + 2 sin x / 2 = x + sin x
dx 2 2 2 2 cos 2 x / 2 cos x / 2 1 + cos x
Hence the result.
∫ ∫(4) Integrals of the form eaxsinbxdx, eaxcosbx dx :
∫ ∫Working rule : To evaluate eax sin bx dx or eax cos bx dx, proceed as follows
(i) Put the given integral equal to I.
(ii) Integrate by parts, taking eax as the first function.
(iii) Again, integrate by parts taking eax as the first function. This will involve I.
(iv) Transpose and collect terms involving I and then obtain the value of I.
eax sin bx dx . Then I = e ax ax cos bx ax − cos bx
b b
∫ ∫ ∫Let I = I . .
. sin bx dx = −e − ae dx
II
−1 e ax . cos bx a eax .cos bx dx −1 eax .cos bx a eax . sin bx ae ax . sin bx dx
b b b b b
I II
∫ ∫= + = + b −
−1 e ax . cos bx a eax .sin bx a2 eax .sin bx dx −1 eax. cos bx a eax . sin bx a2 I
∫= b + b2 − b2 = b + b2 − b2
I + I. a2 = e ax (a sin bx − b cos bx) ⇒ I = e ax (a sin bx − b cos bx) + c
b2 b2 a2 + b2
∫Thus, e ax sin bx = e ax (a sin bx − b cos bx) + c = e ax sin(bx − tan −1 b ) + c
a2 + b2 a2 + b2 a
∫Similarly eax .cos bx dx = e ax (a cos bx + b sin bx) + c = e ax b2 cos bx − tan−1 b + c
a2 + b2 a2 + a
∫Note : eax .sin(bx + c)dx = e ax [a sin(bx + c) − b cos(bx + c)] + k = e ax sin(bx + c) − tan −1 b +k
a2 + b2 a2 + a
b2
∫ eax. cos(bx + c) dx = eax [a cos(bx + c) + b sin(bx + c)] + k = e ax cos(bx + c) − tan−1 b + k
a2 + b2 a2 + b2 a
Important Tips
x eax e ax
a2 + b2 (a2 + b2)2
∫ [ ]•
x eax sin(bx + c)dx = [a sin(bx + c) − b cos(bx + c)] − (a 2 − b2)sin bx(bx + c) − 2ab cos bx(bx + c) +k
∫• x . eax cos(bx + c) dx = x . eax [a cos(bx + c) + b sin(bx + c)] − (a 2 e ax [(a 2 − b 2 ) cos(bx + c) + 2ab sin(bx + c)] + k
a2 + b2 + b2)2
∫• ax .sin(bx + c)dx = ax + b2 [(log a)sin(bx + c) − b cos(bx + c)] + k
(log a)2
∫• ax .cos(bx + c)dx = ax + b2 [(log a)cos(bx + c) + b sin(bx + c)] + k
(log a)2
Example: 32 ∫If ex sin x dx = 1 e x . a + c, then a = [MP PET 1989,99; Rajasthan PET 1985; IIT 1978]
Solution: (a) 2
Example: 33
Solution: (c) (a) sin x − cos x (b) cos x − sin x (c) tan x + c (d) None of these
∫ e x sin x dx ex (1. sin x − 1. cos x) + c = ex (sin x − cos x) + c
= 12 + 12 2
Clearly a = (sin x − cos x)
∫ ∫If u = e ax cos bx dx and v = e ax sin bx dx, then (a 2 + b 2 )(u2 + v 2 ) =
(a) 2e ax (b) (a 2 + b 2 )e 2ax (c) e 2ax (d) (a 2 − b 2 )e 2ax
e ax cos bx dx e ax sin bx a e ax .bx dx e ax sin bx a v
b b b b
∫ ∫u = = − = −
⇒ bu + av = e ax sin bx ......(i)
Similarly, bv − au = −e ax cos bx .....(ii)
Squaring (i) and (ii) and adding. We get, (a 2 + b 2 ) (u 2 + v 2 ) = e 2ax .
Evaluation of the Various forms of Integrals by use of Standard Results .
∫(1) Integral of the form ax 2 dx +c, where ax 2 + bx +c can not be resolved into factors.
+ bx
∫(2) Integral of the form ax px +q c dx.
2+ bx +
∫(3) Integral of the form dx .
ax 2 + bx + c
∫(4) Integral of the form px + q dx.
ax 2 + bx + c
∫(5) Integral of the form ax 2 f(x) dx, where f(x) is a polynomial of degree 2 or greater than 2.
+ bx + c
∫(6) Integral of the form (i) x4 x 2 + 1 + 1 dx, ∫(ii) x4 x 2 − 1 + 1 dx, where k is any constant
+ x + x
k 2 k 2
∫(7) Integral of the form ax 2 + bx + c dx ∫(8) Integral of the form (px + q) ax 2 + bx + c dx
∫(9) Integral of the form dx
PQ
∫(1) Integrals of the form ax 2 dx , where ax 2 + bx +c can not be resolved into factors.
+ bx + c
a x 2 b c a x b 2 b2 c a b 2 b2 − 4ac
a a 2a 4a2 a 2a 4a2
We have, ax 2 + bx + c = + .x + = + − − = x + −
Case (i) : When b2 − 4ac > 0
dx 1 dx , ∫form dx
ax 2 + bx + c = a x 2 − a 2
∫ ∫∴ x b 2 b 2 − 4ac 2
2a 2a
+ −
1 1 x + b − b 2 − 4ac 1 log 2ax + b − b 2 − 4ac + c
a 2a 2a + c = b 2 − 4ac 2ax + b + b 2 − 4ac
= . log
b 2 − 4ac b b 2 − 4ac
2. 2a x + 2a + 2a
Case (ii) : When b2 − 4ac < 0
∫ ∫dx 1 dx , ∫form dx
ax 2 + bx + c = a x 2 + a 2
x b 2 4ac − b2 2
2a 2a
+ +
x b
+ 2a
1 . 1 tan−1 +c = 2 −1 2ax + b
= a 4ac − b2 tan 4ac − b2 + c
4ac − b2 4ac − b2
2a 2a
∫Working rule for evaluating dx : To evaluate this form of integrals proceed as follows :
ax 2 + bx + c
(i) Make the coefficient of x 2 unity by taking ‘a’ common from ax 2 + bx + c.
(ii) Express the terms containing x 2 and x in the form of a perfect square by adding and subtracting the square
of half of the coefficient of x.
(iii) Put the linear expression in x equal to t and express the integrals in terms of t.
dx or dx or dx standard form. After using
x2 + a2 x2 − a2 a2 − x2
∫ ∫ ∫(iv) The resultant integrand will be either in
the standard formulae, express the results in terms of x.
Example: 34 ∫ dx [Rajasthan PET 1997]
2x2 + x +1 =
(a) 1 tan−1 4 x + 1 + c (b) 1 tan−1 4 x + 1 + c (c) 1 tan−1 4x + 1 + c (d) None of these
7 7 27 7 2 7
Solution: (d) ∫I = 1 dx ∫⇒ I = 1 dx
2 2
x2 + x + 1 x 1 2 7 2
2 2 + 4 + 4
I 1 . 4 . tan −1 x + 1 c I= 2 tan−1 4 x + 1 c
2 7 4 7 7
= + ⇒ +
( 7 / 4)
Example: 35 ∫ sin 2 x cos x x + 5 dx equals
+ 4 sin
(a) tan−1(sin x) + c (b) tan−1(sin x + 2) + c (c) tan−1(sin x + 1) + c (d) None of these
Solution: (b) cos x cos x
+ 4 sin x + 2)2
∫ ∫I = dx = dx
sin 2 x x + 5 (sin +1
Put sin x + 2 = t ⇒ cos x dx = dt
∫I = dt = tan−1 t + c = tan−1(sin x + 2) + c .
t2 +1
∫(2) Integral of the form ax px +q c dx : The integration of the function px + q is effected by
2+ bx + ax 2 + bx + c
breaking px + q into two parts such that one part is the differential coefficient of the denominator and the other
part is a constant.
If M and N are two constants, then we express px + q as px + q = M d (ax 2 + bx + c) + N = M.(2ax + b) + N
dx
= (2aM)x + Mb + N .
Comparing the coefficients of x and constant terms on both sides, we have, p = 2aM ⇒ M = p and
2a
q = Mb + N ⇒ N = q − Mb = q − p b .
2a
px +q p (2ax + b) + q − p b
ax 2 + bx + 2a c 2a dx
∫ ∫Thus, M and N are known. Hence, the given integral is dx =
c ax 2 + bx +
p 2ax + b dx q p b dx p log | ax 2 + bx + c |+ q − p b dx C
2a ax 2 + bx + 2a + bx + c 2a 2a + bx
∫ ∫ ∫= + − = ax 2 +
c ax 2 + c
The integral on R.H.S. can be evaluated by the method discussed in previous section.
∫(i) If b 2 − 4ac < 0, then px +q c dx = p log |ax2 + bx + c|+ (2aq − bp) tan−1 2ax + b + k
ax2 + bx + 2a a 4ac − b2 4ac − b2
∫(ii) If b2 − 4ac > 0, then px +q dx = p log | ax 2 + bx + c|+ (2aq − bp) log 2ax + b − b 2 − 4ac +k
2+ bx + 2a 2a b 2 − 4ac 2ax + b + b 2 − 4ac
ax c
Example: 36 ∫ 2 sin 2θ − cosθ dθ =
6 − cos2 θ − 4 sinθ
(a) 2 log |sin 2 θ − 4 sinθ + 5|+7 tan−1(sinθ − 2) + c (b) 2 log |sin 2 θ − 4 sinθ + 5|−7 tan−1(sinθ − 2) + c
(c) − 2 log |sin 2 θ − 4 sinθ + 5|+7 tan−1(sinθ − 2) + c (d) − 2 log |sin 2 θ − 4 sinθ + 5|−7 tan−1(sinθ − 2) + c
Solution: (a) 2(2 sinθ cosθ ) − cosθ dθ (4 sinθ − 1) cosθ
6 − (1 − sin 2 θ ) − 4 sinθ sin 2 θ − 4 sinθ + 5
∫ ∫I = ⇒ I= dθ
∫Put sinθ = t ⇒ cosθ dθ = dt , ∴ I = t 2 4t − 1 5 dt
− 4t +
Let 4t −1 = M d (t 2 − 4t + 5) + N ⇒ 4t − 1 = M(2t − 4) + N
dt
∫Comparing the coefficient of t and constant terms on both side, then M = 2, N = 7 ∴I = 2(2t − 4) + 7 dt
t2 − 4t + 5
I=2 2t − 4 dt 7dt ⇒ I = 2 log | t 2 − 4t + 5| +7 dt
− 4t + t2 − 4t + 5 (t − 2)2 + 1
∫ ∫ ∫⇒ t 2 5 +
⇒ I = 2 log | t2 − 4t + 5| +7 tan−1(t − 2) + c = 2 log |sin 2 θ − 4 sinθ + 5| +7 tan−1(sinθ − 2) + c
∫(3) Integral of the form dx : To evaluate this form of integrals proceed as follows :
ax 2 + bx + c
(i) Make the coefficient of x 2 unity by taking a common from ax 2 + bx + c.
dx 1 dx .
∫ ∫Then,
= a x2 + b x + c
ax 2 + bx + c a a
(ii) Put x2 + b x + c , by the method of completing the square in the form, A2 − X 2 or X 2 + A2 or
a a
X 2 − A2 where, X is a linear function of x and A is a constant.
(iii) After this, use any of the following standard formulae according to the case under consideration
dx = log | x + x 2 + a2 |+c and dx = log | x + x 2 − a2 |+c.
x2 + a2 x2 − a2
dx = sin −1 x + c ⇒
a
∫ ∫ ∫a2 − x2
∫Note : If a < 0, b2 − 4ac > 0, then dx 1 a sin−1 2ax + b + k.
= − b2 − 4ac
ax 2 + bx + c
∫ If a > 0, b2 − 4ac < 0, then dx 1 sinh−1 2ax + b + k
= a 4ac − b2
ax2 + bx + c
∫ If a > 0, b2 − 4ac > 0 dx 1 cosh−1 2ax + b + k
= a b2 − 4ac
ax 2 + bx + c
Example: 37 ∫ dx equals [EAMCET 1996]
2 − 3x − x2
(a) tan −1 2x + 3 + c (b) sec −1 2 x+3 + c (c) sin −1 2x + 3 + c (d) cos −1 2x + 3 + c
17 17 17 17
Solution: (c) dx dx
=
∫ ∫I =
2 − 3x − x2 17 2 3 2
2 x 2
− +
Put x + 3 / 2 = t ⇒ dx = dt
∫I = dx = sin−1 t + c = sin −1 2x + 3 + c
17 17
17 2 t2 / 2
2
−
Example: 38 ∫ dx
=
x2 − 4x + 2
(a) log | x − 2 + x 2 + 2 − 4 x |+c (b) log | x − 2 − x 2 + 2 − 4 x |+c
(c) log | x − 2 + x2 + 2 + 4 x |+c (d) log | x − 2 − x 2 + 2 + 4 x |+c
Solution: (a) ∫ ∫I = dx ⇒ I = dt
(x − 2)2 − 2 t 2 − ( 2)2
Put x − 2 = t ⇒ dx = dt ⇒ I = log | t + t 2 − 2 | +c ⇒ I = log | x − 2 + x 2 − 4 x + 2 | +c
∫(4) Integral of the form px + q dx : To evaluate this form of integrals, first we write,
ax 2 + bx + c
px + q = M d (ax 2 + bx + c) + N ⇒ px + q = M(2ax + b) + N
dx
Where M and N are constants.
By equating the coefficients of x and constant terms on both sides, we get
p = 2aM ⇒ M = p and q = bM + N ⇒N = q − bp
2a 2a
In this way, the integral breaks up into two parts given by
∫ ∫ ∫px + q c dx = p 2ax +b c dx + q − bp dx = I1 + I 2 , (say)
ax 2 + bx 2a ax 2 + bx + 2a ax 2 + bx + c
+
∫Now, I1 = p 2ax + b dx
2a ax 2 + bx + c
ax 2 we have, p t −1 / 2dt p t 1/ 2 p ax 2 + bx + c + C1
∫Putting +c dt, I1 2a 2a . + C1 a
+ bx ⇒ (2ax + b)dx = = = 1 =
2
and I 2 is calculated as in the previous section.
px + q dx = p 2aq − bp dx .
ax 2 + bx + c a 2a ax 2 + bx + c
∫Note : ax 2 + bx + c +
Example: 39 ∫ 5 − 2x dx =
6 + x − x2
(a) 2 6 + x − x2 − 4 sin−1 2x − 1 + c (b) 2 6 + x − x2 + 4 sin−1 2x − 1 + c
5 5
(c) −2 6+ x − x2 −4 sin−1 2x − 1 + c (d) −2 6+ x − x2 +4 sin−1 2x − 1 + c
5 5
Solution: (b) ∫I = 5 − 2x dx
6+ x − x2
Let 5 − 2x = M d (6 + x− x2)+ N ⇒ 5 − 2x = M(1 − 2x) + N
dx
Equating the coefficients of x and constant terms on both sides, we get
−2 = −2M ⇒ M = 1 and 5 = M + N ⇒ N = 5 − 1 = 4 ∴ 5 − 2x = (1 − 2x) + 4
∫ ∫ ∫Hence, I = (1 − 2x) + 4 dx = dx
6+ x − x2 6+ x − x2
1 − 2x dx + 4 = I1 + 4I 2 , (say)
6+ x − x2
∫Now, I1 = 1 − 2x dx
6+ x − x2
Putting 6 + x − x 2 = t ⇒ (1 − 2x)dx = dt, we have,
dt 6 + x − x2 + C1 and dx dx
t = 6 − (x2 − x)
6 + x − x2
∫ ∫ ∫I1 =
=2 t + C1 = 2 I2 =
∫ ∫ ∫I = dx = dx = du 1 1 2 2 where, u = x − 1
4 4 2
6 − x 2 − x + + 25 x 1 5 u2
4 − − 2 2 −
= sin −1 u + C2 = sin −1 2 x − 1 + C2 = sin −1 2x − 1 + C2
5/2 5 2 5
∴ I = I1 + 4I2 = 2 6+ x− x2 + 4 sin −1 2x − 1 + C (where, C = C1 + 4C2)
5
Example: 40 ∫ x+2 equals
x2 − 2x + 4
(a) x2 − 2x + 4 + 3 sinh −1 (x − 1) + c (b) x2 − 2x + 4 − 3 sinh −1 (x − 1) + c
3 3
(c) x2 − 2x + 4 + 3 cosh −1 (x − 1) + c (d) x2 − 2x + 4 − 3 cosh −1 (x − 1) + c
3 3
x+2 x − 1 + 3 dx =
(x − 1)2 + 3 2
x + 2 dx =
∫ ∫ ( ) ∫ ( ) ∫ ∫x2 − 2x + 4
Solution: (a) (x − 1)2 + 3 2 dx = x − 1 dx + 3 dx
x2 − 2x + 4 (x − 1)2 + ( 3) 2
Put x 2 − 2x + 4 = t 2 in the first expression ⇒ 2(x − 1)dx = 2tdt
⇒ (x − 1)dx = tdt
∫ ∫ ∫x + 2 tdt dx
x 2 − 2x + 4 dx = t + 3 (x − 1)2 + ( 3)2
∫ x + 2 dx = x2 − 2x + 4 + 3 sinh −1 x − 1 + c
x 2 − 2x + 4
3
∫(5) Integrals of the form ax 2 f(x) + c dx, where f(x) is a polynomial of degree 2 or greater than 2:
+ bx
To evaluate the integrals of the above form, divide the numerator by the denominator. Then, the integrals take
the form given by ax 2 f(x) +c = Q(x) + ax 2 R(x) + c dx
+ bx + bx
where, Q(x) is a polynomial and R(x) is a linear polynomial in x.
∫ ∫ ∫Then, we have f(x) R(x)
ax 2 + bx + dx = Q(x)dx + ax 2 + bx + c dx
c
The integrals on R.H.S. can be obtained by the methods discussed earlier.
Example: 41 (x 3 + 8)(x − 1) dx equals. [Rajasthan PET 1989]
Solution: (a) x2 − 2x + 4
Example: 42
(a) x3 + x2 − 2x + c (b) x 3 + x 2 − 2x + c (c) (x 3 + x2 − x) + c (d) None of these
Solution: (c) 3 2 3
∫ ∫ ∫ ∫(x3
x2
+ 8)(x − 1) dx = (x + 2)(x 2 − 2x + 4)(x − 1) = (x + 2)(x − 1)dx = (x 2 + x − 2)dx = x3 + x2 − 2x + c
− 2x + 4 x2 − 2x + 4 3 2
∫The value of 2x 3 − 3x2 + 5x + 6 dx is
x2 + 3x + 2
(a) x2 + 3x + 4 ln| x 2 + 3x + 2| +12 ln x +1 + c (b) x + 3x 2 + 4 ln| x + 1|−12ln(x + 2) + c
x +2
(c) (x 2 + 3x) + 8 ln | x +1 +c (d) None of these
(x + 2)2
∫ 2x 3 − 3x2 + 5x + 6 dx
x2 + 3x + 2
(2x + 3) − 8x (2x + 3) dx dx
+ 3x x2 + 3x + 2 (x + 1)(x + 2)
∫ ∫ ∫ ∫= =
x2 + 2 dx (2x + 3) − 4 + 12
∫ ∫= (x2 + 3x) − 4 ln| x2 + 3x + 2|+12 dx − 12 dx
x +1 x+2
= (x 2 + 3x) − 4 ln(x + 1) − 4 ln(x + 2) + 12ln x +1 +c
x+2
= x 2 − 3x + 8 ln(x + 1) − 16 ln(x + 2) + c = (x2 − 3x) + 8 ln | x + 1| +c.
(x + 2) 2
x2 +1 x2 −1
+ kx2 + + kx2 + 1
∫ ∫(6) Integrals of the form
x4 1 dx and x4 dx : To evaluate the integral of the form
∫I = x4 x2 +1 1 dx , proceed as follows
+ kx 2 +
∫(i) Divide the numerator and denominator by x 2 to get I = 1+ 1
x2
1 dx.
x2
x2 + k +
(ii) Put x − 1 = t ⇒ 1 + 1 dx = dt and x2 + 1 −2 = t2 ⇒ x2 + 1 = t2 + 2.
x x2 x2 x2
∫Then, the given integral reduces to the form I = t2 dt , which can be integrand as usual.
+2+k
∫(iii) To evaluate I= x4 x2 −1 + 1dx, we divide the numerator and denominator by x2 and get
+ kx 2
1− 1
x2
∫I = 1 dx
x2 x2
+ k +
Then, we put x + 1 = t ⇒ 1 − 1 dx = dt and x2 + 1 + 2 = t2 ⇒ x2 + 1 = t 2 − 2.
x x2 x2 x2
∫Thus, we have t = t2 dt + k, which can be evaluated as usual.
−2
Important Tips
2x 2 x2 + 1 dx x2 − 1 dx
x4 + x4 +1 x4 +1
∫ ∫ ∫• Algebraic twins: 1 dx = +
2 dx x2 + 1 dx x2 − 1 dx , 2x 2 dx, 2 dx
4+ x4 + 1 x4 + 1 +1+k 1+
∫ ∫ ∫ ∫ ∫x1 = − x4 x2 (x4 + kx2)
x2 + 1 dx x2 − 1 x 2 dx I1 + I2
x4 +1 x4 + 1 x4 +1 2
∫ ∫ ∫ ∫We know the result of I1 = 4 +
and I2 = dx, so for x 1 dx and for , we can use the result of
and I1 − I2 .
2
dx dx ± sin x ± cos x
x + cos 4 x + cos 6 a + b sin x cos
∫ ∫ ∫ ∫ ∫• Trigonometric twins: , , dx
tan x dx, cot x dx, (sin 4 x) sin 6 x
Example: 43 ∫ x2 −1 1 dx =
x4 + x2 +
(a) 1 log x2 + x + 1 + c (b) 1 log x2 − x − 1 + c (c) log x 2 − x + 1 + c (d) 1 log x2 − x + 1 + c
2 x2 − x + 2 x2 + x + 1 x 2 + x + 1 2 x2 + x + 1
1
Solution: (d) ∫ 1− 1 1− 1 1 x + 1 −1 1 x2 +1− x 1 x2 − x +1
x2 x2 2 x + x +1 2 x2 +1+ x 2 x2 + x +1
1 dx ∫= 2 dx = log 1 + c = log +c = log +c
x2 1 x
x2 +1+ x + x − 1
Indefinite integral
Example: 44 ∫ tan x dx =
Solution: (a)
(a) 1 tan−1 tan x − 1 + 1 log tan x − 2 tan x + 1 + c
2 2 tan x 22 tan x + 2 tan x + 1
(b) 1 tan−1 tan x − 1 − 1 log tan x − 2 tan x +1 +c
2 2 tan x 22 tan x + 2 tan x +1
(c) log tan x − 1 + c
2 tan x
(d) None of these
∫Let I = tan x dx
Putting tan x = t ⇒ tan x = t 2 , we have sec 2 xdx = 2tdt ⇒ (1 + tan 2 x)dx = 2tdt
i.e., (1 + t 4 )dx = 2tdt ⇒ dx = 2t dt
1+ t4
tan xdx = t. 2t 2t 2 (t 2 + 1) + (t 2 − 1)dt t2 +1 t2 −1
+t 1+ t4 t4 +1 t4 +1 t4 +1
∫ ∫ ∫ ∫ ∫ ∫∴ I = 1 4 dt = dt = = dt + dt
1+ 1 1 − 1 1 1 1 1
t2 t2 − + t2 + − t2
dt + dt dt dt .
∫ ∫ ∫ ∫= 1 1 = 1 2 + 1 2
t2 t2 t t
t2 + t2 + t + 2 t − 2
Again putting t − 1 = u ⇒ 1 + 1 dt = du in the first integral and t + 1 = v ⇒ 1 − 1 dt = dv in the second
t t2 t t2
integral, we have,
du dv 1 tan −1 u 1 log v − 2 +c
u2 + ( 2)2 + v2 − ( 2)2 2 2 22 v + 2
∫ ∫I = = +
= 1 t 1 + 1 log t + 1 − 2 = 1 tan−1 t2 −1 + 1 log t2 − t 2 +1 +c
2 tan−1 −t 22 t + t + +c 2 t 2 22 t2 + t 2 +1
2 1
t 2
= 1 tan−1 tan x −1 + 1 log tan x − 2 tan x +1 +c
2 2 tan x 22 tan x + 2 tan x +1
Example: 45 ∫ ( )tan x + cot x dx [IIT 1989; Roorkee 1984; Karnataka CET 1991; Pb. CET 1997]
Solution: (a)
(a) 2 tan−1 tan x − cot x + c (b) 2 tan−1 tan x + cot x + c
2 2
(c) tan−1 tan x + cot x + c (d) None of these
2
∫ ( )I = tan x + cot x dx
Put tan x = t2 ⇒ sec 2 x dx = 2t dt ⇒ dx = 2t dt
1+ t4
t 1 2t 1 + 1 1 + 1 dt
t +t t2 t2
∫ ∫ ∫ ( )I = + . dt =2 dt =2
1 4 t2 1 1 2 22
+ t2 +2− 2 t
t − +
1 1 + 1 dt dp 2 tan−1 p t − 1
t t2 p2 + 2 2 = 2 2 t
∫ ( )Putt− = p ⇒ = dp = +c = 2 tan−1 + c
2
= 2 tan−1 tan x − cot x + c
2
∫(7) Integrals of the forms ax 2 + bx + c dx : To evaluate this form of integrals, express ax 2 + bx + c in
[ ]the form a (x + α)2 + β 2 by the method of completing the square and apply the standard result discussed in the
above section according to the case as may be.
dx = (2ax + b) ax 2 + bx +c 4ac − b 2 dx
4a 8a ax 2 + bx + c
∫ ∫Note :
ax 2 + bx + c +
Example: 46 ∫ x2 + 8x + 12 dx [CBSE 1999]
Solution: (b) (a) 1 (x + 4) x 2 + 8x + 12 + 2 log| x + 4 + x 2 + 8x + 12 |+ c
Example: 47 2
Solution: (a)
(b) 1 (x + 4) x 2 + 8x + 12 − 2 log| x + 4 + x 2 + 8x + 12 |+ c
2
(c) (x + 4) x 2 + 8x + 12 + log| x + 4 + x 2 + 8x + 12 |+ c
(d) (x + 4) x 2 + 8x + 12 − log| x + 4 + x 2 + 8x + 12 |+ c
∫ ∫Let I = x2 + 8x + 12 dx = (x2 + 8x + 16) − 4 dx
∫ ∫ ∫= (x + 4)2 − 22 dx = t 2 − 22 dx = t 2 − 22 dt. (putting x + 4 = t ⇒ dx = dt)
= 1 t t2 − 22 − 1 .22 log | t + t2 − 22 |+c = 1 (x + 4) (x + 4)2 − 4 − 2 log | x + 4 + (x + 4)2 − 4 | +c
2 2 2
= 1 (x + 4) x2 + 8x + 12 − 2 log | x + 4 + x2 + 8x + 12 | +c .
2
∫ 2ax − x 2 dx =
(a) 1 (x − a) 2ax − x 2 + 1 a2 sin −1 x − a + c (b) 1 (x − a) 2ax − x 2 − 1 a 2 sin −1 x − a + c
2 2 a 2 2 a
(c) 1 (x − a) 2ax − x 2 + 1 a2 cos −1 x − a + c (d) 1 (x − a) 2ax − x 2 − 1 a2 cos −1 x − a + c
2 2 a 2 2 a
∫ ∫ ∫ ∫2ax − x2 dx = a2 − a2 + 2ax − x2 dx = a2 − (x2 − 2ax + a2) dx = a2 − (x − a)2 dx
= 1 (x − a) 2ax − x2 + 1 a2 sin−1 (x − a) + c
2 2 a
∫(8) Integrals of the form (px + q) ax 2 + bx + c dx : To evaluate this form of integral, proceed as follows:
(i) First express (px + q) as px + q = M d (ax 2 + bx + c) + N ⇒ px + q = M(2ax + b) + N
dx
Where, M and N are constant.
(ii) Compare the coefficients of x and constant terms on both sides, will get
p = 2a M ⇒ M = p and q = Mb + N ⇒ N = q − Mb = q − p b.
2a 2a
(iii) Now, write the given integral as
ax 2 + bx +c dx p (2ax + b) ax 2 bx c dx q p b ax 2 + bx + c dx
∫ ∫ ∫(px + q) = 2a + + + − 2a
= p I1 + q − p b I 2 , (say).
2a 2a
(iv) To evaluate I1, put ax 2 + bx + c = t and to evaluate I2, follows the method discussed in (7)
Example: 48 ∫(2x + 3) x2 + 4 x + 3 dx =
Solution: (d)
(a) log |(x + 2) + ( x 2 + 4 x + 3) | +c (b) log |(x + 2) + ( x 2 + 4 x + 3) | +c
Example: 49
Solution: (d) (c) log |(x − 2) + ( x 2 + 4 x + 3) | +c (d) None of these
Let 2x + 3 = M d (x 2 + 4 x + 3) + N ⇒ 2x + 3 = M(2x + 4) + N
dx
Equating the coefficients of x and constant terms on both sides, we get
2 = 2M ⇒ M = 1 and 3 = 4M + N ⇒ N = 3 − 4 × 1 = −1
∴ 2x + 3 = (2x + 4) − 1
∫ ∫ ∫Hence, I = [(2x + 4) − 1] x2 + 4 x + 3dx = (2x + 4) x2 + 4 x + 3 dx − x2 + 4 x + 3 dx = I1 − I 2 ,(say)
∫Now, I1 = (2x + 4) x2 + 4 x + 3 dx
Putting x 2 + 4 x + 3 = t ⇒ (2x + 4)dx = dt, we have
t 1 / 2dt t3/2 c1 2 2 3)3 / 2 x 2 + 4 x + 3 dx = (x + 2)2 − 12 dx
∫ ∫ ∫I1 = = 3/2 + = 3 (x + 4x + + c1 I2 =
= 1 (x + 2) (x + 2)2 − 12 − 1 .12 log x+2+ (x + 2)2 − 12 + c2 = 1 (x + 2) x2 + 4x + 3 − 1 log x + 2+ x2 + 4x + 3 + c2
2 2 2 2
∴ I = I1 − I2 = 2 (x 2 + 4x + 3)3 / 2 − 1 (x + 2) x2 + 4x + 3 − 1 log x + 2+ x2 + 4x + 3 + c, (where, c = c1 − c 2 ),
3 2 2
∫ (2x − 5) x2 − 4 x + 3 dx = [CBSE 1995]
(a) log | x − 2 + x 2 − 4 x + 3 | +c (b) log | x − 2 − x 2 − 4 x + 3 | +c
(c) log | x + 2 + x 2 − 4 x + 3 | +c (d) None of these
Let 2x − 5 = M d (x2 − 4x + 3) + N ⇒ 2x − 5 = M(2x − 4) + N
dx
Equating the coefficients of x and constant terms on both sides, we get
2= 2M ⇒ M =1
−5 = −4 M + N ⇒ N = 4 M − 5 = −1
∫ ∫ ∫Hence, I = {(2x − 4) − 1} x2 − 4 x + 3 dx ⇒ I = (2x − 4) x2 − 4 x + 3 dx − x2 − 4 x + 3 dx I = I1 − I2 (say)
(2x − 4) x2 − 4 x + 3 dx 2 (x2 3)3 / 2 I2 = x2 − 4 x + 3 dx = x2 − 4 x + 4 − 4 + 3 dx
∫ ∫ ∫I1 = = 3 − 4x + + c1
∫= (x − 2)2 − 12 ⇒ I2 = 1 (x − 2) x2 − 4x + 3 −1 . 12 log (x − 2) + x2 − 4 x + 3
2 2
= 1 (x − 2) x2 − 4x + 3 −1 log (x − 2) + x 2 − 4 x + 3 + c2 , Therefore I = I1 − I2
2 2
I = 2 (x 2 − 4x + 3)3 / 2 + 1 (−2) x2 − 4x + 3 − 1 log (x − 2) + x 2 − 4 x + 3 + c (Where c = c1 + c2)
3 2 2
∫(9) Integrals of the form dx , (where P and Q and linear or quadratic expressions in x): To
PQ
evaluate such types of integrals, we have following substitutions according to the nature of expressions of P and Q
in x :
(i) When Q is linear and P is linear or quadratic, we put Q = t 2 .
(ii) When P is linear and Q is quadratic, we put P = 1 .
t
(iii) When both P and Q are quadratic, we put x = 1 .
t
Example: 50 ∫ dx [Pb. CET 1991]
=
(x − 3) x + 1
(a) 1 log x + 1 + 2 + c (b) 1 log x +1 + 2 + c (c) 1 log x + 1 − 2 + c (d) None of these
2 x + 1 − 2 2 x +1 − 2 2 x + 1 + 2
Solution: (c) Put x + 1 = t 2 ⇒ dx = 2t dt
∫ ∫dx 2t dt ( ) x + 1 = t 2 ⇒ x = t 2 − 1 ⇒ x − 3 = t 2 − 4
= (t 2 − 4)t
(x − 3) x + 1
∫= 2 dt = 2. 1 log t − 2 +c = 1 log x + 1 − 2 + c .
t2 − 22 2.2 t + 2 2 x + 1 + 2
dx dx .
a + b cosx a + b sin x
∫ ∫Integrals of the form
and
To evaluate such form of integrals, proceed as follows:
(1) Put cos x 1 − tan2 x and sin x 2 tan x .
1 + tan2 2 + 2
= x = x
2 tan2 2
1
(2) Replace 1+ tan 2 x in the numerator by sec 2 x .
2 2
(3) Put tan x =t so that 1 sec 2 x dx = dt.
2 2 2
∫(4) Now, evaluate the integral obtained which will be of the form dt by the method discussed earlier.
at 2 + bt + c
∫(i) dx
a + b cos x
∫Case I: When a > b , then dx = 2 tan −1 a − b tan x + c
a + b cos x a2 − b2 a + b 2
∫Case II: When a < b , then dx 1 log b − a tan x + b+a
a + b cos x b2 − a2 2 +c
= x
2 b+a
b − a tan −
∫Case III: When a = b , then a + dx x = 1 tan x + c.
b cos a 2
∫(ii) dx
a + b sin x
∫Case I: When a 2 > b2 or a > 0 and a > b , then dx 2 −1 a tan x +b
a + b sin x a2 − b2 a2 2 b2
= tan + c
−
∫Case II: When a 2 < b 2 , then dx 1 (a tan x + b) − ( b2 − a2 )
a + b sin x = b2 − 2 +c
log x
a2 2 b2 − a2
(a tan + b) +
Case III: When a 2 = b 2
In this case, either b = a or b = −a
∫(a) When b = a , then a+ dx = −1 cot π + x + c = 1 [tan x − sec x] + c
b sin x a 4 2 a
∫(b) When b = −a , then a + dx = 1 tan π + x + c.
b sin x a 4 2
Example: 51 ∫ dx = [Rajasthan PET 1997]
3 + 4 cos x
Solution: (b)
Example: 52 (a) 1 log tan(x / 2) − 7 + c (b) 1 log tan(x / 2) + 7 + c
7 tan(x / 2) + 7 7 tan(x / 2) − 7
(c) 1 log 7 + tan(x / 2) + c (d) 1 log 7 − tan(x / 2) + c
7 7 − tan(x / 2) 7 7 + tan(x / 2)
dx 1 b − a tan x + b + a 1 tan x + 7
a + b cos x = 2 7 log tan 2 −
∫If b > a then, log x + c ⇒ I= x +c
b2 − a2 2 2
b − a tan − b + a 7
∫If dx x = tan x + a + b, then [Roorkee 1979]
1 + sin 2
(a) a = π , b=3 (b) a = −π , b=3
4 4
(c) a = π , b = arbitrary constant (d) a= −π , b = arbitrary constant
4 4
Solution: (d) ∫If a = b , then dx x = − 1 cot π + x + c
Example: 53 a + b sin a 4 2
∫∴ dx = − cot π + x = − tan π − π − x + c = − tan π − x + c = tan x − π + c
1 + sin x 4 2 2 4 2 4 2 2 4
Hence a = −π and b = arbitrary constant.
4
∫ dx [Roorkee1983; Rajasthan PET 1990, 97]
5 + 4 cos x =
(a) 2 tan −1 1 tan x + c (b) 1 tan −1 1 tan x + c (c) 2 tan −1 1 tan x + c (d) 1 tan −1 1 tan x + c
3 3 3 3 3 3 2 3 3 2
Solution: (c) ∫If a > b, then dx 2 . tan −1 a −b tan x +c
a + b cos x = a2 − b2 a +b 2
∫∴ dx = 2 tan −1 1 tan x + c .
4 cos x 3 3 2
5+
dx dx
cos x + a sin x + bcos x
∫ ∫Integrals of the form b c x , .
a + sin
∫(1) Integral of the form a + dx + csinx : To evaluate such integrals, we put b = r cosα and
b cos x and
c = r sinα.
r2 = b2 + c2 = tan−1 c dx dx
b cos x a + r cos(x − α)
∫ ∫So that, and
α . ∴I= a + r(cosα + sinα sin x) =
∫Again, Put x − α = t ⇒ dx = dt, we have I = dt
a + r cos t
Which can be evaluated by the method discussed earlier.
∫(2) Integral of the form dx : To evaluate this type of integrals we substitute a = r cosθ ,
a sin x + b cos x
b = r sinθ and so r = a2 + b2 , α = tan −1 b
a
dx 1 dx 1
a sin x + b cos x = r sin(x + α) = r
∫ ∫ ∫So, cos ec(x + α)dx
= 1 log tan x + α = 1 log tan x + 1 tan −1 b +c
r 2 2 a2 + b2 2 2 a
Note : The integral of the above form can be evaluated by using cos x = 1 − tan2 x / 2
1 + tan2 x / 2
sin x = 1 2 tan x / 2 .
+ tan2 x /2
Important Tips
If a > b, a2 > b2 + c 2, then dx 2 tan−1 (a − b) tan x / 2 + c k
∫• a + b cos x + c sin = +
x a2 − b2 − c2 a2 − b2 − c2
If a > b, a2 < b2 + c 2, then dx 1 log (a − b) tan x / 2 + c − b2 + c2 − a2 k
∫• a + b cos x + c sin x = +
b2 + c2 − a2 (a − b) tan x / 2 + c + b2 + c 2 − a2
If a < b, dx −1 log (b − a) tan x / 2 − c − b2 + c2 − a2 k.
∫• a + b cos x + c sin x = +
b2 + c2 − a2 (b − a)tan x / 2 − c + b2 + c 2 − a2
Example: 54 ∫ dx = [BIT Ranhi 1990; Rajasthan PET 1990, 97, 99; Karnataka CET 1999]
sin x + cos x
(b) log tan(π / 8 − x / 2) + c
(a) log tan(π / 8 + x / 2) + c
(c) 1 log tan(π / 8 + x / 2) + c (d) None of these
2
Solution: (c) ∫ ∫dx ⇒ sec2 x / 2 dx
2 tan x / 2 1 − tan2 x / 2 2 tan x / 2 + 1 − tan2 x / 2
1+ tan2 x /2 + 1 + tan2 x / 2
Put tan x / 2 = t ⇒ 1 sec 2 x/2 dx = dt
2
I=2 dt =2 dt
2t + 1 − t 2 2 − (t 2 − 2t + 1)
∫ ∫∴
I=2 dt 2 log 2 +t −1 +c ⇒ I = 1 log [( 2 − 1) + tan x / 2][ 2 − 1] + c
∫⇒ ( 2)2 − (t − 1)2 = 22 2 −t +1 2 [( 2 + 1) − tan x / 2][ 2 − 1]
⇒ I= 1 log tan π / 8 + tan x / 2 + 1 log ( 2 − 1) + c ⇒ I= 1 log tan(π / 8 + x / 2) + c1
2 1 − ( 2 − 1) tan x / 2 2 2
where c1 = 1 log ( 2 − 1) + c
2
∫ ∫ ∫Trick : I = 1 1 1 = 1 log tan (π / 8 + x / 2) + c .
2 2 2 2
1 dx = dx = cosec (π / 4 + x) dx
sin x + 1 cos x sin(π / 4 + x)
22
Example: 55 ∫ dx = [EAMCET 2002]
Solution: (c) 1 − cos x − sin x
Example: 56 (a) log 1 + cot x +c (b) log 1 − tan x +c (c) log 1 − cot x +c (d) log 1 + tan x +c
Solution: (b) 2 2 2 2
∫Given I = dx
1 − cos x − sin x
dx sec2 x / 2 . dx
∫ ∫I = ⇒ I= 1 + tan2 x / 2 − 1 + tan2 x / 2 − 2 tan x / 2
(1 − tan 2 x / 2) 2 tan x /2
1 − (1 + tan 2 x / 2) − 1 + tan 2 x/2
sec2 x / 2 . dx 1 / 2. sec 2 x / 2.dx
2 tan2 x / 2 − 2 tan x / 2 ⇒ tan 2 x / 2 − tan x / 2
∫ ∫I =
1 2 therefore I = dt dt − 1 1 dt dt
2 t2 −t t (t − 1) ⇒ t − 1 −1 t
∫ ∫ ∫ ∫ ∫Put tan x / 2 = t
⇒ sec x / 2.dx = dt ⇒I= + t dt ⇒ I = t −
I = log(t − 1) − log t + c ⇒ I = log t −1 +c ⇒ I = log tan x / 2 − 1 +c ⇒ I = log 1 − cot x / 2 + c
t tan x / 2
The antiderivative of f(x) = 3 + 5 sin 1 whose graph passes through the point (0,0) is
x + 3 cos x
(a) 1 log 1− 5 tan x / 2 (b) 1 log 1+ 5 tan x / 2 (c) 1 log 1+ 5 cot x / 2 (d) None of these
5 3 5 3 5 3
dx sec 2 x / 2 dx 1 1 log 5 tan x / 2 1 +c
(3 + 5 sin x + 3 cos x) = 10 tan x / 2 + 6 5 5 3
∫ ∫y = = log(5 tan x / 2 + 3) + c = +
Passes through (0,0)
∴c =0 then y = 1 log 1+ 5 tan x / 2 .
5 3
dx , dx , dx , dx dx
bcos2 bsin2 + bcos2 + bcosx)2 a + bsin2x + ccos2x
∫ ∫ ∫ ∫ ∫Integrals of the form a x a x 2 x x ,
+ + asin (asinx
To evaluate the above forms of integrals proceed as follows:
(1) Divide both the numerator and denominator by cos2 x.
(2) Replace sec 2 x in the denominator, if any by (1 + tan2 x).
(3) Put tan x = t ⇒ sec 2 xdx = dt.
(4) Now, evaluate the integral thus obtained, by the method discussed earlier.
Example: 57 ∫ dx [Roorkee 1989; CBSE PMT 1992; DCE 2001]
Solution: (b) 1 + 3 sin 2 x =
Example: 58 (a) 1 tan −1 (3 tan 2 x) + c (b) 1 tan −1 (2 tan x) + c (c) tan−1(tan x) + c (d) None of these
Solution: (c) 3 2
Example: 59
Solution: (c) ∫ ∫sec2 x dx sec2 x dx
sec2 x + 3 tan2 x = 1 + 4 tan2 x
Put 2 tan x = t
sec 2 x dx = dt
2
∫∴I = 1 dt = 1 tan −1 t + c ⇒ I = 1 tan−1(2 tan x) + c .
2 1+ t2 2 2
∫ dx [AISSE 1986; CBSE PMT 1992]
4 sin 2 x + 5 cos 2 x
(a) 1 tan−1 2 tan x + c (b) 1 tan−1 tan x + c (c) 1 tan−1 2 tan x + c (d) None of these
5 5 5 5 25 5
∫ sec 2 x dx Put 2 tan x = t
4 tan 2 x + 5 2 sec2 x dx = dt
∫⇒ 1 dt = 1 tan −1 2 tan x + c
2 + ( 5)2 25 5
t2
∫ dx [MP PET 1994]
(2 sin x + cos x)2
(a) 1 2 1 + 1 (b) 1 log(2 tan x + 1) + c (c) 1 x +c (d) − 1 2 1 − 1 + c
2 tan x 2 2 + cot 2 tan x
∫ ∫ ∫dx cosec 2 x dx
(2 sin x + cos x)2 = (2 + cot x)2
dx =
sin 2 x(2 + cot x)2
∫Put (2 + cot x) = t ⇒ − cosec 2 x dx = dt = −dt = 1 +c = 1 + c
t2 t 2 + cot x
Example: 60 ∫ cos x dx =
Solution: (a) cos 3x
Example: 61 (a) 1 log 1 + 3 tan x + c (b) − 1 log 1 + 3 tan x + c
Solution: (b) 2 3 1 − 3 tan x 2 3 1 − 3 tan x
(c) 1 tan−1( 3 tan x) + c (d) None of these
6
cos x cos x
cos 3x cos3 x − 3
∫ ∫Let I = dx = 4 cos x dx
∫Dividing the numerator and denominator by cos2 x, we have I = dx
4 cos 2 x − 3
dt 1 dt 1 dt 1 . 1 log 1 +t +c = 1 log 1 + 3t +c
1 − 3t2 = 3 3 2. 1 3 −t 23 1− 3t
∫ ∫ ∫I = 1 = 3 1 2 = 1
3 − t2 3 3
− t2 3
I = 1 log 1 + 3 tan x + c.
2 3 1 − 3 tan x
∫ dx equals
4 sin 2 x + 4 sin x .cos x + 5 cos 2 x
(a) tan−1 tan x + 1 + c (b) 1 tan−1 tan x + 1 + c (c) 4 tan−1 tan x + 1 + c (d) None of these
2 4 2 2
∫I = dx ∫ sec 2 x dx ∫= 1 sec 2 x dx
4 sin 2 x + 4 sin x .cos x + 5 cos 2 x 4
= 4 tan 2 x + 5 + 4 tan x tan x + 1 2 +1
2
1 sec 2 1 dt 1 tan −1 1 tan−1 1
∫ ∫Put 2 =t x dx dt 4 2+ 4 4 tan x 2 c
tan x + ⇒ = = t 1 = t + c ⇒ I = + + .
a sinx + b cosx a sinx + bcos x + q
c sinx + d cosx c sinx + d cosx + r
∫ ∫Integrals of the form
and
∫(1) Integrals of the form a sinx + b cos x dx : Such rational functions of sin x and cos x may be
c sinx + d cos x
integrated by expressing the numerator of the integrand as follows:
Numerator = M (Diff. of denominator)+N (Denominator)
i.e., a sin x + b cos x = M d (c sin x + d cos x) + N(c sin x+d cos x)
dx
The arbitary constants M and N are determined by comparing the coefficients of sin x and cos x from two sides
of the above identity. Then, the given integral is
a sin x + b cos x M(c cos x − d sin x) + N(c sin x + d cos x) dx c cos x − d sin x
c sin x + d cos x c sin x + d cos x c sin x + d cos x
∫ ∫ ∫ ∫I = dx = =M dx + N 1dx
= M log |c sin x + d cos x |+ Nx + c.
∫(2) Integrals of the form a sinx + bcosx + q dx : To evaluate this type of integrals, we express the
csinx + dcosx + r
numerator as follows: Numerator = M(Denominator) + N(Differentiation of denominator) + P
i.e.,(c sin x + b cos x + q) = M(c sin x + d cos x + r) + N(c cos x − d sin x) + P.
where M, N, P are constants to be determined by comparing the coefficients of sin x,cos x and constant term
on both sides.
a sin x + b cos x + q dx M dx + N Diff.of denominator dx dx
c sin x + d cos x + r Denominator c sin x − d cos x + r
∫ ∫ ∫ ∫∴ = +
∫= Mx + N log | Denominator |+P c sin x dx x + r .
+ d cos
Indefinite integral
Important Tips
∫• a cos x + b sin x dx = ac + bd x + ad − bc log|c cos x + d sin x | +c .
c cos x + d sin x c2 + d2 c2 + d2
Example: 62 ∫3 cos x + 3 sin x dx = [EAMCET 1991]
Solution: (a) 4 sin x + 5 cos x
Example: 63 (a) 27 x − 3 log(4 sin x + 5 cos x) + c (b) 27 x + 3 log(4 sin x + 5 cos x) + c
Solution: (b) 41 41 41 41
(c) 27 x − 3 log(4 sin x − 5 cos x) + c (d) None of these
41 41
3 cos x + 3 sin x = M d (4 sin x + 5 cos x) + N(4 sin x + 5 cos x)
dx
⇒ 3 cos x + 3 sin x = M(4 cos x − 5 sin x) + N(4 sin x + 5 cos x)
⇒ Comparing the coefficient of sin x and cos x on both sides.
⇒ −5M + 4 N = 3 and 4M + 5N = 3 ⇒ M = −3 , N = 27
41 41
−3 (4 cos x − 5 sin x) + 27 (4 sin x + 5 cos x) 27 −3 4 cos x − 5 sin x
41 4 sin x + 41 41 41 4 sin x + 5 cos x
dx I= dx dx
∫ ∫ ∫∴ I = ⇒ +
5 cos x
⇒ I = 27 x − 3 log(4 sin x + 5 cos x) + c .
41 41
∫ dx x = [CBSE 1992; PB 1993; HB 1994]
1 + cot
(a) 1 x + 1 log | sin x + cos x | +c (b) 1 x − 1 log | sin x + cos x | +c
2 2 2 2
(c) −1 x + 1 log | sin x + cos x | +c (d) None of these
2 2
dx dx sin x
1 + cot x = x+ cos
∫ ∫ ∫Here, I = = dx
cos x sin x
1 + sin x
Let sin x = M d (sin x + cos x) + N(sin x + cos x) ⇒ sin x = M(cos x − sin x) + N(sin x + cos x)
dx
Comparing the coefficients of sin x and cos x of both the sides, we have 1 = −M + N and 0 = M + N
Solving these equations, we have M = −1 and N = 1
2 2
∴ sin x = − 1 (cos x − sin x) + 1 (sin x + cos x)
2 2
−1 (cos x − sin x) + 1 (sin x + cos x) 1 cos x − sin x 1 1 1
2 sin 2 =−2 sin x + cos x 2 2 2
∫ ∫ ∫Hence, I = dx dx + 1 dx = − log | sin x + cos x|+ x + c.
x + cos x
Integration of Rational Functions by using Partial Fractions .
(1) Proper rational functions: Functions of the form f(x) , where f(x) and g(x) are polynomial and
g(x)
g(x) ≠ 0, are called rational functions of x.
If degree of f(x) is less than degree of g(x), then f(x) is called a proper rational function.
g(x)
(2) Improper rational function : If degree of f(x) is greater than or equal to degree of g(x), then f(x) , is
g(x)
called an improper rational function and every improper rational function can be transformed to a proper rational
function by dividing the numerator by the denominator.
For example, x3 is an improper rational function and can be expressed as (x + 5) + 19x − 30 ,
x2 − 5x + 6 x2 − 5x + 6
which is the sum of a polynomial (x + 5) and a proper function 19x − 30 .
x2 − 5x + 6
(3) Partial fractions: Any proper rational function can be broken up into a group of different rational
fractions, each having a simple factor of the denominator of the original rational function. Each such fraction is
called a partial fraction.
If by some process, we can break a given rational function f(x) into different fractions, whose denominators
g(x)
are the factors of g(x), then the process of obtaining them is called the resolution or decomposition of f(x) into its
g(x)
partial fractions.
Depending on the nature of the factors of the denominator, the following cases arise.
Case I: When the denominator consists of non-repeated linear factors: To each linear factor (x − a)
occurring once in the denominator of a proper fraction, there corresponds a single partial fraction of the form
x A a , where A is a constant to be determined.
−
Case II: When the denominator consists of linear factors, some repeated: To each linear factor
(x − a) occurring r times in the denominator of a proper rational function, there corresponds a sum of r partial
fractions of the form.
A1 + (x A2 + ........... + (x Ar
x−a − a)2 − a)r
Where A' s are constants to be determined. Of course, Ar is not equal to zero.
Case III: When the denominator consists of quadratic factors: To each irreducible non repeated
quadratic factor ax 2 + bx + c, there corresponds a partial fraction of the form ax Ax +B c , where A and B are
2+ bx +
constants to be determined.
To each irreducible quadratic factor ax 2 + bx + c occurring r times in the denominator of a proper rational
fraction there corresponds a sum of r partial fractions of the form
A1 x + B1 + A2 x + B2 + ................. + Ar x + Br
ax 2 + bx + c (ax 2 + bx + c)2 (ax 2 + bx + c)r
Where, A's and B's are constants to be determined.
(4) General methods of finding the constants
(i) In the given proper fraction, first of all factorize the denominator.
(ii) Express the given proper fraction into its partial fractions according to rules given above and multiply both
the sides by the denominator of the given fraction.
(iii) Equate the coefficients of like powers of x in the resulting identity and solve the equations so obtained
simultaneously to find the various constant is short method. Sometimes, we substitute particular values of the
variable x in the identity obtained after clearing of fractions to find some or all the constants. For non-repeated
linear factors, the values of x used as those for which the denominator of the corresponding partial fractions become
zero.
Note : If the given fraction is improper, then before finding partial fractions, the given fraction must be
expressed as sum of a polynomial and a proper fraction by division.
(5) Special cases: Some times a suitable substitution transform the given function to a rational fraction which
can be integrated by breaking it into partial fractions.
Example: 64 ∫ (1 + cos x sin x) dx = [Roorkee 1979]
Solution: (c) sin x)(2 +
Example: 65
Solution: (b) (a) log [(1 + sin x)(2 + sin x)] + c (b) log 2 + sin x + c
1 + sin x
Example: 66
Solution: (a) (c) log 1 + sin x + c (d) None of these
2 + sin x
Put sin x = t ⇒ cos x dx = dt
∫ ∫ ∫cos x dx 1 1 t 1
(1 + sin x)(2 + sin x) = +1 + t 2
dt − t 2 dt = log(t + 1) − log(t + 2) + c = log + + c
(1 + t)(2 + t) = t +
= log 1 + sin x + c.
2 + sin x
∫ (x 3x + 1 2) = [Pb. CET 1996]
− 2)2 (x +
(a) 5 log x + 2 + 7 2) + c (b) 5 log x − 2 + 7 2) + c
16 x − 2 4(x − 16 x + 2 4(x −
(c) 16 log x − 2 − 7 2) + c (d) None of these
5 x + 2 4(x −
We have, 3x + 1 = (x A 2) + B + C
(x − 2)2(x + 2) − (x − 2)2 (x + 2)
3x + 1 = A(x − 2)(x + 2) + B(x + 2) + C(x − 2)2 .......(i)
Putting x = 2and - 2 successively in equation (i), we get B= 7 , C= −5
4 16
Now, we put x =0 and get A= 5
16
3x +1 dx 5 dx 7 dx −5 dx 5 log(x 2) 7 1 5 log(x 2) + c
− 2)2(x + 16 x−2 4 (x − 2)2 16 x+2 16 4 − 16
∫ ∫ ∫ ∫(x 2) = + + = − − (x 2) − +
= 5 log x−2 − 7 2) + c
16 x+2 4(x −
∫ x2 + 1 dx = [MP PET 1999]
x(x 2 − 1)
(a) log x2 −1 +c (b) − log x2 − 1 + c (c) log x + c (d) − log x + c
x x x2 + x2 +
1 1
∫I = 2x 1 − 1 dx ⇒ I = log(x 2 − 1) − log x + c ⇒ I = log x2 −1 + c .
x2 − x x
Example: 67 ∫If (x 2 2x 2 + 3 dx = a log x − 1 + b tan −1 x + c then values of a and b are [Rajasthan PET 2000]
Solution: (d) − 1)(x 2 + 4) x + 1 2
(d) 1 / 2, 1 / 2
Example: 68 (a) 1, − 1 (b) −1, 1 (c) 1 / 2, − 1 / 2
Solution: (a)
Example: 69 Put x2 = y
Solution: (b) ∴ (x2 2x2 + 3 4) = (y 2y + 3 4) ⇒ (y 2y + 3 4) = (y A + (y B
− 1)(x2 + − 1)(y + − 1)(y + − 1) + 4)
∴ 2y + 3 = A(y + 4) + B(y − 1)
Comparing the coefficient of y and constant terms
⇒ A + B = 2 and 4 A − B = 3
∴ A = 1 and B = 1
I= 1 dx 1 dx ⇒ I= 1 dx 1 dx I 1 log x −1 1 tan −1 x +c
∫ ∫ ∫ ∫∴ − + + 2− + + ⇒ = 2 x +1 + 2 2
y 1 y 4 x 1 x 2 4
∴ a= 1 and b= 1 .
2 2
∫ (x − x4 + 1) dx = [Roorkee 1986]
1) (x 2
(a) x(x + 2) + log (x − 1) − log (x2 + 1) − tan−1 x + c (b) x(x + 2) + log (x − 1) + log (x2 + 1) − tan−1 x + c
2 2 4 2 2 2 4 2
(c) x(x + 2) + log (x − 1) + log (x2 + 1) + tan−1 x + c (d) None of these
2 2 4 2
x4 dx x4 − 1 dx 1 dx (x + 1)(x − 1)(x 2+ 1) dx dx
1)(x − 1)(x 1) (x 2 (x − 1)(x2 + 1) (x − 1)(x2 + 1)
∫ ∫ ∫ ∫ ∫(x− 2 + 1) = (x 2 + 1) + (x − + 1) = +
(x + 1) dx + dx x2 x 1 log(x − 1) 1 log(x 2 1) 1 tan−1 x c
∫ ∫= (x − 1)(x2 + 1) = 2 + + 2 − 4 + − 2 +
= x(x + 2) + log(x − 1) − log(x 2 + 1) − tan−1 x + c.
2 2 4 2
∫ tanφ + tan3 φ dφ
1+ tan3 φ
(a) 1 log |1 + tanφ |+ 1 log | tan2 φ − tanφ + 1| + 1 tan−1 2 tanφ − 1 + c
3 6 3 3
(b) −1 log |1 + tan φ |+ 1 log | tan2 φ − tan φ + 1|+ 1 tan−1 2 tan φ − 1 + c
3 6 3 3
(c) 1 log 1 + tan2 φ + c
3 tan3 φ
(d) None of these
tan φ + tan3 φ tan φ(1 + tan2 φ) tan φ sec2 φ
1+ tan3 φ 1 + tan3 φ 1 + tan3 φ
∫ ∫ ∫Let I = dφ = ⇒ I= dφ
t t dt
1+ t3 (1 + t)(1 − t + t 2 )
∫ ∫Putting tanφ = t ⇒ sec 2φ dφ = dt, we have, I = dt =
Let t = A + Bt+C
(1 + t)(1 − t + t 2) 1+ t 1− t + t2
⇒ t = A(1 − t + t 2 ) + (Bt + C)(1 + t) ......(i)
Putting t = −1 in equation (i), we get, −1= A(1 + 1 + 1) ⇒ A = − 1
3
Comparing the coefficients of t 2 and constant terms on both the sides of equation (i) we get
0 = A + B ⇒ B = −A = 1 and 0= A + C ⇒ C = −A = 1
3 3
∴ (1 + t t + t2) = − 1 . 1 + 1 . t +1
t)(1 − 3 1+r 3 1− t + t2
1 dt 1 t +1 1 dt 1 2t + 2 1 dt 1 (2t − 1) + 3
3 1+ t 3 t2 − t +1 =−3 1+t + 6 2− t + 3 1+ t 6 t2 − t +1
∫ ∫ ∫ ∫ ∫ ∫Hence,I = − + dt t 1 dt = − + dt
∫ ∫ ∫1 2t −1 1 dt
t2 − t +1 2
=−3
dt 1 3 dt dt 1
1+t + 6 6 t2 − t +1 1+t + 6
∫ ∫ ∫1 2t −1 dt + . dt + 1 2 3
t2 − t +1 2 4
=−3 t − +
1 1 1 2 t 1 1 1 1 tan−1 2t − 1
3 6 2 3 −2 3 6 3 3
= − log |1 + t | + log | t2 − t + 1| + . tan−1 + c = − log |1 + t |+ log |t2 − t + 1|+ + c
3
2
= − 1 log |1 + tanφ |+ 1 log | tan2 φ − tanφ + 1| + 1 tan−1 2 tanφ − 1 + c.
3 6 3 3
Example: 70 ∫ dx [Roorkee 1979; Pb. CET 1991; Rajasthan PET 1998; Karnataka CET 1996]
Solution: (a) x(x n + 1) =
(a) 1 log xn + c (b) n log xn + 1 + c (c) −1 log xn + c (d) − n log xn + 1 + c
n xn + xn n xn + xn
1 1
dx x n −1
x(xn + 1) = n(xn +
∫ ∫Let I = x 1)dx
Putting x n = t ⇒ nx n−1dx = dt, we have
1 dt 1 1 1 dt, (by resolving into partial fractions)
n t(t + 1) = n t +
∫ ∫I= − t 1
= 1 [log t − log (t + 1)] + c = 1 log t +c = 1 log xn + c.
n n t +1 n xn +1
Integration of Trigonometric Functions .
∫(1) Integral of the form sinm x cosn x dx : (i) To evaluate the integrals of the form
∫I = sinm x cosn x dx, where m and n are rational numbers.
(a) Substitute sin x = t, if n is odd;
(b) Substitute cos x = t, if m is odd;
(c) Substitute tan x = t, if m + n is a negative even integer; and
(d) Substitute cot x = t, if 1 (n − 1) is an integer.
2
(e) If m and n are rational numbers and m + n − 2 is a negative integer, then substitution cos x = t or
2
tan x = t is found suitable.
∫(ii) Integrals of the form R(sin x,cos x)dx, where R is a rational function of sin x and cos x, are transformed
into integrals of a rational function by the substitution tan x = t, where −π < x < π. This is the so called universal
2
substitution. Sometimes it is more convenient to make the substitution cot x =t for 0< x< 2π .
2
The above substitution enables us to integrate any function of the form R(sin x,cos x). However, in practice, it
sometimes leads to extremely complex rational function. In some cases, the integral can be simplified by:
∫(a) Substituting sin x = t, if the integral is of the form R(sin x)cos x dx.
∫(b) Substituting cos x = t, if the integral is of the form R(cos x)sin x dx.
(c) Substituting tan x = t, i.e., dx = 1 dt 2 , if the integral is dependent only on tan x.
+t
(d) Substituting cos x = t , if R(− sin x, cos x) = −R(sin x, cos x)
(e) Substituting sin x = t , if R(sin x,− cos x) = −R(sin x, cos x)
(f) Substituting tan x = t , if R(− sin x,− cos x) = −R(sin x, cos x)
Important Tips
∫ ∫ ∫ ∫• To evaluate integrals of the form sin mx cos nx dx, sin mx .sin nx dx, cos mx .cos nx dx and cos mx .sin nx dx, we use the following
trigonometrical identities.
sin mx .cos nx = 1 [sin(m − n)x + sin(m + n)x] ⇒ cos mx .sin nx = 1 [sin(m + n)x − sin(m − n)x]
2 2
sin mx .sin nx = 1 [cos(m − n)x − cos(m + n)x] ⇒ cos mx .cos nx = 1 [cos(m − n)x + cos(m + n)x]
2 2
Example: 71 ∫ sin3 x cos2x dx [Pb. CET 1992; EAMCET 1996]
Solution: (a)
Example: 72 (a) cos2 x − cos3 x + c (b) cos5 + cos3 +c (c) sin5 − sin3 +c (d) sin5 + sin3 +c
Solution: (c) 5 3 5 3 5 3 5 3
Example: 73
Solution: (d) ∫I = sin x(1 − cos2 x) cos2 x dx
∫Put cos x = t (t 2 − t4 ) dt t5 t3 cos5 x cos3 x c.
⇒ − sin x dx = dt ⇒ I=− = 5 − 3 + c = 5 − 3 +
∫ dθ =
sinθ .cos3 θ
(a) log tanθ + tan 2 θ + c (b) log tan θ − 1 tan2 θ + c (c) log tan θ + 1 tan2 θ + c (d)None of these
2 2
sec 2 θ dθ sec 2 θ (1 + tan 2 θ )dθ
sinθ cosθ tanθ dθ
∫ ∫ ∫dθ = =
sinθ cos3 θ
1+ t2 1 t2 1
t t 2 2
∫ ∫Put t = tanθ ⇒ dt = sec 2 θ dθ then it reduces to ⇒ dt = + t dt = log t + +c = log tanθ + tan 2 θ + c.
∫ sin3 2x dx = [Karnataka CET 1999]
cos 5 2x
(a) tan4 x + c (b) tan 4 x + c (c) tan4 2x + x + c (d) 1 tan 4 2x + c
8
sin 3 2x . The given equation may be written as sin 3 2x 1 tan3 2x . sec 2 2x dx.
∫ ∫ ∫Given, I = cos 5 2x dx cos 3 2x . cos 2 2x dx =
Put tan 2x = t and 2 sec 2 2x dx = dt
∫I = 1 t3 dt = t4 + c = tan4 2x + c.
2 8 8
Example: 74 ∫If In = sin nx dx, where n > 1, then In − In−2 =
Solution: (b) sin x
(a) (n 2 1) cos(n − 1)x (b) n 2 1 sin(n − 1)x (c) 2 cos n x (d) 2 sin n x
− − n n
∫In = sin nx dx
sin x
sin(n − 2)x In − In−2 sin nx − sin(n − 2)x dx 2 cos(n − 1)x .sin x
sin x sin x sin x
∫ ∫ ∫In−2 = dx ⇒ = = dx
In − I n− 2 = 2 sin(n − 1)x
(n − 1)
(2) Reduction formulae for special cases
sinn − cos x . sinn−1 x n−1 sinn−2 x dx
n n
∫ ∫(i) x dx = +
cosn sin x cosn−1 x n−1 cosn−2 x dx
n n
∫ ∫(ii) x dx = +
tann tann−1 x tann−2 x dx
n−1
∫ ∫(iii) x dx = −
cot n −1 n −1 cotn−2 x dx
∫ ∫(iv) x dx = n−1 cot x −
∫ [ ∫ ](v) 1
sec n x dx = (n − 1) sec n− 2 x. tan x + (n − 2) secn−2 x dx
∫ [ ∫ ](vi) 1
cosecn x dx = (n − 1) − cos ecn−2 x . cot x + (n − 2) cosecn−2 x dx
sin p x cosq sinq+1 x. cosp−1 x p −1 sinp−2 x. cosq x dx
− p+q p+q
∫ ∫(vii)
x dx = +
sin p x cosq sinp+1 x . cosq−1 x p −1 sinp x. cosq−2 x dx
p+q p+q
∫ ∫(viii)
x dx = +
dx x (2n − 3) dx
(x 2 + k)n = k(2n − 2)(x 2 k(2n − 2) (x 2 + k)n−1
∫ ∫(ix) +
+ k)n−1
Important Tips
Reduction formulae for I(n,m) = sin n x dx is I (n,m) 1 . sin n−1 x (n − 1) . I (n− 2, m− 2)
∫• cos m x = m− 1 cos m−1 x − (m − 1)
Example: 75 ∫ tan5 θ dθ = [EAMCET 1993]
Solution: (c)
(a) tan4 θ − tan 2 θ + log secθ +c (b) tan4 θ − tan 2 θ − log secθ +c
4 2 4 2
(c) tan4 θ − tan 2 θ + log |secθ |+c (d) None of these
4 2
∫ tan5 θ dθ = I5 = tan4 θ − I3 = tan4 θ − tan 2 θ + I1 = tan4 θ − tan 2 θ + log |secθ |+c.
4 4 2 4 2
Example: 76 ∫ cosec4 x dx [Rajasthan PET 2002]
Solution: (c)
(a) cot x + cot 3 x + c (b) tan x + tan 3 x + c (c) cot x − cot 3 x + c (d) − tan x − tan 3 x + c
3 3 3 3
cosec 2 x(1 + cot 2 x)dx cosec 2 x dx + (cot x)3
∫ ∫ ∫ ∫cosec4x dx = = cot 2 x cosec 2 x dx = − cot x + 3 + c.
Example: 77 ∫Integrate sin8 x dx
Solution: (a)
(a) 1 sin 8x −8 sin 6x + 28 sin 4 x − 56 sin 2x + 35x + c
27 8 6 4 2
(b) 1 − sin 8x +8 sin 6x + 28 sin 4 x + 56 sin 2x − 35x + c
27 8 6 4 2
(c) sin 8x − 8 sin 6x + 28 sin 4 x − 56 sin 2x + 35 x + c
8 6 4 2
(d) None of these
Let cos x + i sin x = y ; then
2 cos x = y + 1 , 2cos nx = yn + 1 ⇒ 2i sin x = y − 1 , 2 i sin nx = yn − 1 (Remember as the standard results)
y yn y yn
Thus 28 i8 sin8 x = y − 1 8 = y8 + 1 − 8 y6 + 1 + 28 y 4 + 1 − 56 y 2 + 1 + 70
y y8 y6 y4 y2
= 2 cos 8x − 16 cos 6x + 56 cos 4 x − 112 cos 2x + 70
Thus sin 8 x = 1 (cos 8x − 8 cos 6x + 28 cos 4 x − 56 cos 3x + 35),
27
∫and sin8 x dx = 1 sin 8x −8 sin 6x + 28 sin 4 x − 56 sin 2x + 35x +c
27 8 6 4 2
Integration of Hyperbolic Functions
∫(1) sinh x dx = cos hx + c ∫(2) cos hx dx = sin hx + c
∫(3) sec h 2 x dx = tan h x + c ∫(4) cosec h2 x dx = − cot hx + c
∫(5) sec hx tan hx dx = − sec hx + c ∫(6) cosec hx cot h dx = −cosec hx + c
Example: 78 ∫ cos−1 1 dx = [Rajasthan PET 2002]
x
Solution: (b)
Example: 79 (a) x sec −1 x + cos h−1 x + c (b) x sec −1 x − cos h−1 x + c
Solution: (a) (c) cos h−1 x − x sec −1 x + c (d) None of these
∫ ∫ ∫cos−1 1 dx 1 x dx = x sec −1 x − cos h−1 x + c.
x x x2 −1
= sec −1 x dx = x sec −1 x −
∫ cos x − sin x dx equals [Rajasthan PET 1996]
sin 2x
(a) cos h−1(sin x + cos x) + c (b) sin h−1(sin x + cos x) + c
(c) − cos h−1(sin x + cos x) + c (d) − sin h−1(sin x + cos x) + c
cos x − sin x dx =
1 + sin 2x − 1
∫ ∫ ∫I = cos x − sin x dx = cos x − sin x dx
sin 2x (cos x + sin x)2 − 1
∫Put cos x + sin x = t ⇒ (cos x − sin x) dx = dt ⇒ dt = cos h−1t + c = cos h−1(sin x + cos x) + c.
t2 −1
Example: 80 ∫ sin h −1 x dx equals [Rajasthan PET 1996]
Solution: (d)
Example: 81 (a) x cos h−1x − x 2 + 1 + c (b) x cos h−1x + x 2 + 1 + c
Solution: (a) (c) x sin h−1x + x 2 + 1 + c (d) x sin h−1x − x 2 + 1 + c
Example: 82
Solution: (c) ∫ ∫sin h−1x .1dx = x sin h−1x − 1 .x dx = x sin h−1x − 1 + x 2 + c.
Example: 83 I II 1+ x2
Solution: (d)
∫ ex + a dx =
Example: 84 ex − a
Solution: (b) [ ](b) 1
(a) cos h−1(e x / a) + sec−1(e x / a) + c a cos h−1(e x / a) + sec −1(e x / a) + c
[ ](c) 1
a cos h−1(e x / a) + sec −1(e x / a) + c (d) None of these
ex + a dx = e x dx +
e2x − a2 e2x − a2
ex ex /a
∫ ∫ ∫ ∫ ∫ ∫ex
+ a dx = a dx = (e x / a)2 − 12 dx + 1 dx
− a e2x − a2 (e x / a)2 − 12
ex ex dt dt = cos h−1(e x / a) + sec−1(e x / a) + c.
a a +
∫ ∫Put = t ⇒ dx = dt =
t2 −1 t t2 −1
∫ dx
(e x + e −x )2 =
(a) tan hx + c (b) 1 tan hx + c (c) 1 tan hx + c (d) None of these
2 4
4 . dx
∫ ∫ ∫dx 1 1 sec h2x dx 1 tan hx c.
(e x + e−x )2 = 4 4
(e x + e−x )2 = 4 = +
∫ tan x dx =
1 − tan 2 x
(a) cos h−1( 2 cos x) + c (b) − cos h−1( 2 cos x) + c (c) 1 cos −1( 2 cos x) + c (d) −1 cos h−1( 2 cos x) + c
22
sin x dx sin x dx
= 2 cos 2 x − 1
cos 2 x − sin 2 x
∫ ∫ ∫tan x dx
=
1 − tan 2 x
Put 2 cos x = t ⇒ − 2 sin x dx = dt
∫I = −1 dt = −1 cos h−1( 2 cos x + c) .
2 t2 −1 2
∫ sin x sin hx dx =
(a) 1 (e x − e−x ) + c (b) 1 [sin x cos hx − cos x sin hx] + c
2 2
(c) 1 [sin x cos hx + cos x cos hx] + c (d) None of these
2
∫ ∫ ∫I = sin x sin hx dx = − sin hx cos x + cos hx cos x dx = sin hx cos x + [cos hx sin x − sin hx sin x dx]
I = 1 [sin x cos x − cos x sin hx] + c.
2
Integral of the type f [x,(ax + b)m1/n1 ,(ax + b)m2/n2 ...] where f is a rational function and
m1, n1, m2, n2 are Integers
To evaluate such type of integral, we transform it into an integral of rational function by putting (ax + b) = t s,
where s is the least common multiple (L.C.M.) of the numbers n1,n2.
∫Integrals of the form xm(a + bxn)p dx
Case I : If p ∈ N (Natural number). We expand the integral with the help of binomial theorem and integrate.
Example: 85 ∫Evaluate x1/ 3 (2 + x1/ 2 )dx
Solution: (d)
(a) 3x4 / 3 + 7 x7 / 3 + c (b) x4 / 3 + 5 x5 / 3 + c (c) 3x4 / 3 + 3 x5 / 3 + c (d) 3x4 / 3 + 3 x7 / 3 + 24 x11 / 6 + c
3 3 5 7 11
∫I = x1/ 3 (2 + x1/ 2 )2 dx = 4x4/3 + x7/3 + 4.x11/ 6 +c = 3x 4 / 3 + 3 x7/3 + 24 x 11 / 6 + c.
4/3 7/3 11 / 6 7 11
Since P is natural number,
∫ ∫I = x1/ 3 (4 + x + 4 x1/ 2 )dx = (4 x1/ 3 + x 4 / 3 + 4 x 5 / 6 )dx
Case II: If p ∈ I − (i.e. negative integer). Write x = t k , where k is the L.C.M. of denominator of m and n.
Example: 86 ∫Evaluate x −2 / 3 (1 + x 2 / 3 )−1 dx
Solution: (a)
(a) 3 tan−1(x1 / 3) + c (b) 3 tan−1 x + c (c) 3 tan−1(x2 / 3) + c (d) None of these
If we substitute x = t 3 (as we know P ∈ negative integer)
∴ Let x = t k , where k is L.C.M. of denominator m and n.
3t 2dt dt 3 tan −1(x1/ 3 ) + c .
2 (1 + t 2 t2 +1
∫ ∫∴ x = t 3 ⇒ dx = 3t 2dt or I = t ) = 3 = 3 tan−1(t) + c ⇒ I =
Case III: If m+1 is an integer and p∈ fraction put (a + bxn) = tk , where k is the denominator of the fraction p.
n
Example: 87 ∫Evaluate x −2 / 3 (1 + x1/ 3 )1/ 2 dx
Solution: (b)
(a) 2(1 + x1 / 3)2 / 3 + c (b) 2(1 + x1 / 3)3 / 2 + c (c) 2(1 + x 2 / 3)3 / 2 + c (d) 2 (1 + x2 / 3)2 / 3 + c
If we substitute 1 = x1/ 3 = t 2 then, 1 dx = dt
3x 2 / 3
t .6t dt 6 t 2 dt = 2t 3 + c or I = 2(1 + x1/ 3 )3 / 2 + c.
1
∫ ∫∴ I = =
Case IV: If m+ 1 + P is an integer and P ∈fraction. We put (a + bxn) = tk . xn, where k is the denominator of
n
the fraction P.
Example: 88 ∫Evaluate x −11(1 + x 4 )−1/ 2dx
(a) 1 (t 5 + t3 + t) + c (b) − 1 t5 − 2t 3 + t + c (c) 1 t4 2t 3 + t + c (d) None of these
2 2 3 +3
5 2 4
where t = 1+ 1
x4
Solution: (b) Here m+ 1 + P = −11 + 1 + 1 = −3
n 4 2
If we substitute then
1 = t2 and −4 ⇒I= dx dx
x4 x5 x11(1 + x 4 )1/ 2 x11. x 2 (1 + 1 / x 4 )1/ 2
∫ ∫1 + dx = 2t dt =
dx 1 2t dt 1 (t 2 − 1)2 dt −1 (t 4 2t 2 + 1)dt −1 t5 2t 3
x13 (1 + 1 / x 4 )1/ 2 = 4 x8t 2 2 3
∫ ∫ ∫ ∫I = = − = − = 2 5 − + t + c,
where t = 1+ 1
x4
Integrals using Euler's substitution
∫Integrals of the form f(x), (ax 2 + bx + c) dx are calculated with the aid of one of the there Euler substitution:
(1) ax 2 + bx + c = t ± x a , if a > 0.
(2) ax 2 + bx + c = t x ± c, if c > 0.
(3) ax 2 + bc + c = (x − α) t, if ax 2 + bx + c = a(x − α)(x − β ), i.e., if x is real root of (ax 2 + bx + c).
Note : The Euler substitution often lead to rather some calculations, therefore they should be applied only
when it is difficult to find another method for calculating the given integral.
Example: 89 ∫Evaluate dx
Solution: (a)
x + x2 − x +1
(a) 2 log e |t|− 1 log e |t − 1|− 3 log e |t + 1|+ (t 3 + c (b) 2 log e |t|+ 1 log e |t − 1|+ 3 log e |t + 1|+ (t 3 + c
2 2 + 1) 2 2 + 1)
(c) 2 log e |t|− 1 log e |t − 1|+ 3 log e |t + 1|+ (t 3 + c (d) None of these
2 2 + 1)
Where t= x 2 − x + 1 + 1
x
Since here c = 1, we can apply the second Euler substitution.
x 2 − x + 1 = tx − 1
Hence (2t − 1)x = (t 2 − 1)x 2 ; x= 2t − 1
t2 −1
∴ dx = 2(t 2 − t + 1)dt and x+ x2 − x +1 = t 1
(t 2 − 1)2 −1
dx − 2t 2 + 2t 1−)22dt
= t(t − 1)(t +
x + x2 − x +1
∫ ∫∴ I =
Using partial fractions, we have,
− 2t 2 + 2t − 2 = A + t B + (t C + (t D or (−2t + 2t − 2) = A(t − 1)(t + 1)2 + B(t + 1)2 + C(t − 1)(t + 1)t + Dt.
t(t − 1)(t + 1)2 t −1 + 1) + 1)2
We get A = 2, B = 1 / 2,C = −3 / 2, D = −3.
dt 1 dt 3 dt dt 1 3 3
t −2 t −1 − 2 + 1) (t + 1)2 2 2 +
∫ ∫ ∫ ∫Hence
I=2 (t − 3 = 2 log e |t | − log e | t − 1|− log e | t + 1| + (t 1) + c
Where t= x 2 − x + 1 + 1
x
Example: 90 ∫Evaluate x dx
Solution: (a) (7x − 10 − x 2)3
(a) −2 −5 + 2t + c (b) 2 −5 − 2t + c (c) −1 5 + 2t + c (d) None of these
9 t 9 t 9 t
where t = 7x − 10 − x 2
x−2
In this case a < 0 and c < 0. Therefore neither (I) nor (II) Euler substitution is applicable. But the quadratic 7x − 10 − x 2
has real roots α = 2, β = 5.
∴We use the (III) i.e., 7x − 10 − x 2 = (x − 2)(5 − x) = (x − 2)t
where (5 − x) = (x − 2)t 2 or 5 + 2t 2 = x(1 + t 2 )
∴ x = 5 + 2t 2 ⇒ (x − 2)t = 5 + 2t 2 − 2 t = 3t , ∴ dx = −6t dt
1+ t2 1+ t2 1+ t2 (1 + t 2 )2
x dx 5 + 2t 2 . (1 − 6t )2 dt −6 5 + 2t 2 −2 5 −2 −5
= 1+ t2 + t2 27 t2 9 t2 9 t
+ 2 dt 2t + c
( 7x − 10 − x 2 )3
∫ ∫Hence, I = 3t 3 ∫= dt = = +
+t
1 2
∫∴ ( x dx = −2 −5 + 2t + c , where t = 7x − 10 − x 2
7x − 10 − x2)3 9 t x−2
Some Integrals which can not be Found
Any function continuous on interval (a, b) has an antiderivative in that interval. In other words, there exists a
function F(x) such that F'(x) = f(x).
However not every antiderivative F(x), even when it exists is expressible in closed form in terms of elementary
functions such as polynomials, trigonometric, logarithmic, exponential etc. function. Then we say that such
antiderivatives or integrals "can not be found." Some typical examples are:
∫(i) dx ∫(ii) e x2 dx ∫(iii) x2 dx
log x 1+ x5
(iv) ∫ 3 1 + x2 dx ∫(v) 1 + x 3 dx (vi) 1 − k 2 sin 2 x dx
∫(vii) e −x2 dx ∫(viii) sin x dx ∫(ix) cos x dx
x x
∫(x) sin x dx ∫(xi) sin(x 2 ) dx ∫(xii) cos(x 2 ) dx ∫(xiii) x tan x dx etc.
Vector-algebra-2
(3) Rotation about an axis : When a rigid body rotates about a fixed axis ON with an angular velocity ω ,
then the velocity v of a particle P is given by v = ω × r , where r = OP and ω =|ω | (unit vector along ON)
N
ω
P
r
O
Example: 42 Three forces i + 2j − 3k, 2i + 3j + 4k and i − j + k are acting on a particle at the point (0, 1, 2). The magnitude of the
Solution: (b)
moment of the forces about the point (1, – 2, 0) is [MNR 1983]
Example: 43
Solution: (b) (a) 2 35 (b) 6 10 (c) 4 17 (d) None of these
Total force F = (i + 2j − 3k) + (2i + 3j + 4k) + (i − j + k) = 4i + 4j + 2k (0,1,2) F
A (1,–2,0)
Moment of the forces about P = r × F = PA × F P
PA = (0 − 1)i + (1 + 2) j + (2 − 0)k = −i + 3j + 2k r
i jk
∴ Moment about P = (−i + 3j + 2k)×(4i + 4j + 2k) = − 1 3 2 = − 2i + 10j − 16k
4 42
Magnitude of the moment = | −2i + 10j − 16k| = 2 12 + 52 + 82 = 2 90 = 6 10
The moment of the couple formed by the forces 5i + k and −5i − k acting at the points (9, − 1, 2) and (3, − 2,1)
respectively is [AMU 1998]
(a) −i + j + 5k (b) i − j − 5k (c) 2i − 2j − 10k (d) −2i + 2j + 10k
Moment of the couple, F 5 i+k
A
= BA × F = {(9 − 3)i + (−1 + 2) j + (2 − 1)k}× (5i + k) (9,–1,2)
i jk (3,–2,1)
= (6i + j + k) × (5i + k) = 6 1 1 = i − j − 5k B
501 –F –5i–k
Scalar Triple Product .
If a,b,c are three vectors, then their scalar triple product is defined as the dot product of two vectors a and b × c . It
is generally denoted by a . (b × c) or [abc]. It is read as box product of a, b, c . Similarly other scalar triple products can
be defined as (b × c).a,(c × a).b . By the property of scalar product of two vectors we can say, a.(b× c) = (a × b).c
(1) Geometrical interpretation of scalar triple product : The scalar triple product of three vectors is equal
to the volume of the parallelopiped whose three coterminous edges are represented by the given vectors. a,b,c
form a right handed system of vectors. Therefore (a × b).c = [abc]= volume of the parallelopiped, whose
coterminous edges are a, b and c.
(2) Properties of scalar triple product
(i) If a,b,c are cyclically permuted, the value of scalar triple product remains the same. i.e.,
(a × b).c = (b × c).a = (c × a).b or [a b c] = [b c a] = [c a b]
(ii) The change of cyclic order of vectors in scalar triple product changes the sign of the scalar triple product
but not the magnitude i.e., [a b c] = −[b a c] = −[c b a] = −[a c b]
(iii) In scalar triple product the positions of dot and cross can be interchanged provided that the cyclic order of
the vectors remains same i.e., (a × b).c = a.(b × c)
(iv) The scalar triple product of three vectors is zero if any two of them are equal.
(v) For any three vectors a, b, c and scalar λ , [λ a b c]= λ[a b c]
(vi) The scalar triple product of three vectors is zero if any two of them are parallel or collinear.
(vii) If a, b, c, d are four vectors, then [(a + b) c d] = [a c d] + [bcd]
(viii) The necessary and sufficient condition for three non-zero non-collinear vectors a,b,c to be coplanar is
that [a b c] = 0 i.e., a,b,c are coplanar ⇔ [a bc] = 0 .
(ix) Four points with position vectors a, b, c and d will be coplanar, if [a b c] + [d c a] + [d a b] = [a b c] .
(3) Scalar triple product in terms of components
(i) If a = a1i + a2j + a3k , b = b1i + b2j + b3k and c = c1i + c2j + c3k be three vectors.
a1 b1 c1
Then, [a b c] = a2 b2 c 2
a3 b3 c3
a1 a2 a3
(ii) If a = a1l + a2m + a3n, b = b1l + b2m + b3n and c = c1l + c2m + c3n , then [a b c] = b1 b2 b3 [l m n]
c1 c2 c3
(iii) For any three vectors a,b and c
(a) [a + b b + c c + a] = 2[a b c] (b) [a − b b − c c − a] = 0 (c) [a × b b × c c × a] = [a b c]2
(4) Tetrahedron : A tetrahedron is a three-dimensional figure formed by four triangle OABC is a tetrahedron
with ∆ABC as the base. OA,OB,OC, AB, BC and CA are known as edges of the A(a)
tetrahedron. OA, BC;OB,CA and OC, AB are known as the pairs of opposite b a
B(b) O
edges. A tetrahedron in which all edges are equal, is called a regular tetrahedron.
Properties of tetrahedron c
(i) If two pairs of opposite edges of a tetrahedron are perpendicular, then
C(c)
the opposite edges of the third pair are also perpendicular to each other.
(ii) In a tetrahedron, the sum of the squares of two opposite edges is the same for each pair.
(iii) Any two opposite edges in a regular tetrahedron are perpendicular.
Volume of a tetrahedron
(i) The volume of a tetrahedron = 1 (area of the base) (corresponding altitude)
3
= 1 . 1 | AB × AC || ED | = 1 | AB × AC || ED |cos 0o for AB × AC|| ED
3 2 6
= 1 (AB × AC). ED = 1 [ AB AC EA + AD] = 1 [ AB AC AD] .
6 6 6
Because AB, AC, EA are coplanar, so [AB AC EA] = 0
(ii) If a,b,c are position vectors of vertices A, B and C with respect to O, then volume of tetrahedron
OABC = 1 [a b c]
6
(iii) If a,b,c,d are position vectors of vertices A, B, C, D of a tetrahedron ABCD, then
its volume = 1 [b − a c − a d − a]
6
(5) Reciprocal system of vectors : Let a,b,c be three non-coplanar vectors, and let
a′ = b×c , b′ = c×a , c′ = a×b .
[abc] [abc] [abc]
a′, b′,c′ are said to form a reciprocal system of vectors for the vectors a,b,c .
If a, b, c and a′, b′,c′ form a reciprocal system of vectors, then
(i) a.a′ = b.b′ = c.c′ = 1 (ii) a.b′ = a.c′ = 0;b.c′ = b.a′ = 0;c.a′ = c.b′ = 0
(iii) [a′ b′c′] = [a 1 c] (v) a,b,c are non-coplanar iff so are a′,b′,c′ .
b
Example: 44 If u, v and w are three non-coplanar vectors, then (u + v − w).[(u − v) × (v − w)] equals [AIEEE 2003]
Solution: (b)
Example: 45 (a) 0 (b) u .(v × w) (c) u .(w × v) (d) 3u .(v × w)
Solution: (c)
(u + v − w).[u − v × (v − w)] = (u + v − w).[(u × v) − (u × w) − 0 + (v × w)]
Example: 46 = [u u v] + [v u v] − [w u v] − [u u w] – [v u w] + [w u w] + [u v w] + [v v w] − [w v w]
Solution: (b) = 0 + 0 − [u v w] − 0 + [u v w] + 0 + [u v w] + 0 − 0 = [u v w] = u.(v × w) .
Example: 47 The value of ‘a’ so that the volume of parallelopiped formed by i + aj + k ; j + ak and ai + k becomes minimum is
Solution: (c)
[IIT Screening 2003]
(a) – 3 (b) 3 (c) 1 / 3 (d) 3
Volume of the parallelepiped
V = [i + aj + k j + ak ai + k] = (i + aj + k) .{(j + ak)× (ai + k)} = (i + aj + k) .{i + a2j − ak)} = 1 + a3 − a
dV = 3a2 − 1 ; d2V = 6a ; dV = 0 ⇒ 3a2 − 1 = 0 ⇒ a = ± 1
da da2 da 3
At a = 1 , d2V = 6 >0
3 da2 3
∴ V is minimum at a = 1
3
If a, b, c be any three non-zero non-coplanar vectors, then any vector r is equal to
(a) z a + xb + yc (b) xa + yb + zc (c) ya + zb + xc (d) None of these
Where x = [r b c] , y= [r c a] , z = [r a b]
[a b c] [a b c] [a b c]
As a, b, c are three non-coplanar vectors, we may assume r = αa + β b + γ c
[r b c] = (α a + β b+γ c).(b × c) = α{a .(b × c)} = α[a b c] ⇒ α = [r b c]
[a b c]
But x= [r b c] ; ∴α = x
[abc]
Similarly β = y, γ = z ; ∴ r = x a + y b + zc .
If a, b, c are non-coplanar vectors and λ is a real number, then the vectors a + 2b + 3c, λb + 4c and (2λ − 1)c are non-
coplanar for [AIEEE 2004]
(a) No value of λ (b) All except one value of λ
(c) All except two values of λ (d) All values of λ
As a, b, c are non-coplanar vectors. ∴ [abc] ≠ 0
Now, a + 2b + 3c, λb + 4c and (2λ − 1)c will be non-coplanar iff (a + 2b + 3c).{(λb + 4c)×(2λ − 1)c} ≠ 0
i.e., (a + 2b + 3c).{λ(2λ − 1)(b × c)} ≠ 0 i.e., λ(2λ − 1)[a b c] ≠ 0
∴ λ ≠ 0, 1
2
Thus, given vectors will be non-coplanar for all values of λ except two values: λ =0 and 1 .
2
Example: 48 x, y, z are distinct scalars such that [xa + yb + zc, xb + yc + za, xc + ya + zb] = 0 where a, b, c are non-coplanar vectors then
(a) x + y + z = 0 (b) xy + yz + zx = 0 (c) x 3 + y3 + z3 = 0 (d) x 2 + y2 + z 2 = 0
a, b, c are non-coplanar
Solution: (a)
∴ [a b c] ≠ 0
Now, [xa + yb + zc, xb + yc + za, xc + ya + zb] = 0
⇒ (xa + yb + zc).{(xb + yc + za)× (xc + ya + zb)} = 0 ⇒ (xa + yb + zc).{(x 2 − yz)(b × c) + (z 2 − xy)(a × b) + (y2 − zx)(c × a)} = 0
⇒ x(x2 − yz)[abc] + y(y2 − zx)[b c a] + z (z2 − xy)[c a b] = 0 ⇒ (x3 − xyz)[a b c] + (y3 − xyz)[abc] + (z3 − xyz)[abc] = 0
As [abc] ≠ 0 , x 3 + y3 + z3 − 3xyz = 0 ⇒ (x + y + z)(x 2 + y2 + z 2 − xy − yz − zx) = 0
⇒ 1 (x + y + z) {(x − y)2 + (y − z)2 + (z − x)2} = 0 ⇒ x+y+z =0 or x=y=z
2
But x, y, z are distinct. ∴ x + y + z = 0 .
Vector Triple Product.
Let a,b,c be any three vectors, then the vectors a × (b × c) and (a × b) × c are called vector triple product of a,b,c .
Thus, a ×(b × c) = (a .c) b − (a . b)c
(1) Properties of vector triple product
(i) The vector triple product a × (b×c) is a linear combination of those two vectors which are within brackets.
(ii) The vector r = a × (b × c) is perpendicular to a and lies in the plane of b and c .
(iii) The formula a × (b × c) = (a .c) b − (a . b)c is true only when the vector outside the bracket is on the left
most side. If it is not, we first shift on left by using the properties of cross product and then apply the same formula.
Thus, (b × c)× a = −{a × (b × c)} = − {(a .c) b − (a . b)c} = (a . b)c − (a .c) b
(iv) If a = a1i + a2j + a3k , b = b1i + b2j + b3k and c = c1i + c2j + c3k
i jk
then a × (b × c) = a1 a2 a3
b2c3 − b3c2 b3c1 − b1c3 b1c2 − b2c1
Note : Vector triple product is a vector quantity.
a × (b × c) ≠ (a × b) × c
Example: 49 Let a, b and c be non-zero vectors such that (a × b)× c = 1 | b || c |a . If θ is the acute angle between the vectors b and
Solution: (a) 3
c , then sinθ equals [AIEEE 2004]
(a) 22 (b) 2 (c) 2 (d) 1
3 3 3 3
(a × b)× c = 13| b||c|a ⇒ (a . c) b − (b . c)a = 1 | b || c |a
3
⇒ (a . c) b = {(b . c) + 1 | b||c|}a ⇒ (a . c) b =| b||c| cosθ + 1 a
3
3
As a and b are not parallel, a.c= 0 and cos θ + 1 =0
3
⇒ cos θ = − 1 ⇒ sin θ = 22
3 3
Example: 50 If a = i + j + k, b = i + j, c = i and (a × b) × c = λa + µ b , then λ + µ = [EAMCT 2003]
(a) 0 (b) 1 (c) 2 (d) 3
(d) a + b + c
Solution: (a) (a × b) × c = λa + µb ⇒ (a . c) b − (b . c)a = λa + µb ⇒ λ = −b . c , µ = a . c
Example: 51
Solution: (c) ∴ λ + µ = a . c − b . c = (a − b). c = {(i + j + k) − (i + j)}.i = k . i = 0 .
If a, b, c and p, q, r are reciprocal system of vectors, then a × p + b × q + c × r equals
(a) [a b c] (b) (p + q + r) (c) 0
p= b × c , q = c×a , r= a×b
[a b c] [a b c] [a b c]
a × p = a × (b × c) = (a . c) b − (a.b)c
[a b c] [a b c]
Similarly b×q = (a.b)c − (b.c)a and c× r = (b . c) a − (a . c) b
[a b c] [a b c]
∴ a×p + b×q + c× r = [a 1 c] {(a . c ) b − (a . b)c + (a . b) c − (b . c) a + (b . c) a − (a . c) b} = [a 1 c] × 0 = 0
b b
Scalar product of Four Vectors.
(a × b).(c× d) is a scalar product of four vectors. It is the dot product of the vectors a × b and c × d .
It is a scalar triple product of the vectors a, b and c× d as well as scalar triple product of the vectors a × b, c and d .
(a × b).(c × d)= a.c a.d
b.c b.d
Vector product of Four Vectors.
(1) (a × b)× (c × d) is a vector product of four vectors.
It is the cross product of the vectors a × b and c × d .
(2) a × {b × (c × d)},{(a × b)× c}× d are also different vector products of four vectors a, b, c and d .
Example: 52 a × [a × (a × b)] is equal to [AMU 2001]
(a) (a × a).(b × a)
(b) a.(b × a) − b.(a × b) (c) [a.(a × b)]a (d) (a.a)(b × a)
Solution: (d) a ×[a × (a × b)] = a ×[(a.b)a − (a.a)b] = (a.b)(a × a) − (a.a) (a × b) = (a . b) 0 + (a . a)(b × a) = (a.a)(b × a)
Example: 53 [b × c c × a a × b] is equal to [MP PET 2004]
(a) a × (b × c) (b) 2[a b c] (c) [a b c]2 (d) [a b c]
Solution: (c) [(b × c)( c × a),(a × b)] = (b × c).{[ c × a)× (a × b)} = (b × c).{[c a b] a − [a a b]c}
= (b × c).{[a b c]a − 0} = [b c a][abc] = [abc]2
Example: 54 Let the vectors a, b, c and d be such that (a × b)× (c × d) = 0 . Let P1 and P2 be planes determined by pair of vectors
a, b and c, d respectively. Then the angle between P1 and P2 is [IIT Screening 2000]
(a) 0o (b) π (c) π (d) π
4 3 2
Solution: (a) (a × b)× (c × d) = 0 ⇒ (a × b) is parallel to (c × d)
Hence plane P1 , determined by vectors a, b is parallel to the plane P2 determined by c, d
∴ Angle between P1 and P2 = 0 (As the planes P1 and P2 are parallel).
Vector Equations.
Solving a vector equation means determining an unknown vector or a number of vectors satisfying the given
conditions. Generally, to solve a vector equation, we express the unknown vector as a linear combination of three
known non-coplanar vectors and then we determine the coefficients from the given conditions.
If a, b are two known non-collinear vectors, then a, b, a × b are three non-coplanar vectors.
Thus, any vector r = xa + yb + z(a × b) where x, y, z are unknown scalars.
Example: 55 If a = i + j + k, a.b = 1 and a × b = j − k , then b = [IIT Screening 2004]
(a) i (b) i − j + k (c) 2j − k (d) 2i
Solution: (a) Let b = b1i + b2 j + b3k
Example: 56 i jk
Solution: (a)
Now, j − k = a × b = 1 1 1 ⇒ b3 − b2 = 0,b1 − b3 = 1,b2 − b1 = −1 ⇒ b3 = b2,b1 = b2 + 1
b1 b2 b3
Now, a.b = 1 ⇒ b1 + b2 + b3 = 1 ⇒ 3b2 + 1 = 1 ⇒ b2 = 0 ⇒ b1 = 1,b3 = 0 . Thus b = i
The point of intersection of r × a = b × a and r × b = a × b where a = i + j and b = 2i − k is [Orissa JEE 2004]
(a) 3i + j − k (b) 3i − k (c) 3i + 2j + k (d) None of these
We have r × b = a × b and r × a = b × a
Adding, r × (a + b) = (a × b) + (b × a)
⇒ r × (a + b) = 0 ⇒ r is parallel to a + b
∴ r = λ(a + b) = λ{(i + j) + (2i − k)} = λ {3i + j − k}
Example: 57 For λ = 1 , r = 3i + j − k [IIT 1995]
Let a = i − j, b = j − k,c = k − i . If dˆ is a unit vector such that a . d = 0 = [b c d] , then dˆ is equal to
(a) ± i + j − k (b) ± i + j + k (c) ± i + j − 2k (d) ±k
3 3 6
Solution: (c) Let dˆ = α i + β j + γ k
a . dˆ = 0 ⇒ (i − j).(α i + βj + γk) = 0 ⇒ α − β = 0 ⇒ α = β
i jk
[b c d] = 0 ⇒ (b × c).d = 0 ⇒ 0 1 − 1 . (α i + β j + γ k) = 0 ⇒ (i + j + k) . (α i + β j + γ k) = 0 ⇒ α + β + γ = 0
−1 0 1
⇒ γ = −(α + β ) = −2α ; (β = α)
|dˆ |= 1 ⇒ α 2 + β 2 + γ 2 = 1 ⇒ α 2 + α 2 + 4α 2 = 1 ⇒ α = ± 1 = β and γ = 2
66
∴ dˆ = ± 1 (i + j − 2k) .
6
Example: 58 Let p, q, r be three mutually perpendicular vectors of the same magnitude. If a vector x satisfies equation
p ×|(x − q)× p| +q×|(x − r)× q | + r ×|(x − p)× r | = 0 , then x is given by [IIT 1997]
(a) 1 (p + q − 2r) (b) 1 (p + q + r) (c) 1 (p + q + r) (d) 1 (2p + q − r)
2 2 3 3
Solution: (b) Let | p| =|q| =| r | = k
∴ p = k pˆ, q = k qˆ, r = k ˆr
Let x = α pˆ + β qˆ + γ ˆr
Now, p × {x − q}× p} = (p.p)(x − q) − {p.(x − q)}p = |p|2 (x − q) − {(p. x) − p.q}p
= k2(x − q) − {| p|(pˆ . xˆ) − 0}| p|pˆ = k 2(x − q) – | p|2 (pˆ . xˆ) pˆ = k2{x − q − α pˆ)
∴ p × {(x − q)× p} + q × (x − r)× q} + r × {(x − p)× r} = 0
⇒ k 2{x − q − αp + x − r − βq + x − p − γr} = 0 ⇒ 3x − (p + q + r) − (αp + βq + γr) = 0
⇒ 3x − (p + q + r) − x = 0 ⇒ 2x − (p + q + r) = 0
∴ x = 1 (p + q + r)
2
Example: 59 Let the unit vectors a and b be perpendicular and the unit vector c be inclined at an angle θ to both a and b. If
Solution: (b)
c = α a + β b + γ (a × b) , then [Orissa JEE 2003]
Example: 60
Solution: (b) (a) α = β = cosθ , γ 2 = cos 2θ (b) α = β = cosθ ,γ 2 = − cos 2θ
(c) α = cosθ , β = sinθ ,γ 2 = cos 2θ (d) None of these
We have, |a|=|b|= 1
a . b = 0 ; (as a ⊥b)
c = α a + βb + γ (a × b) ......(i)
Taking dot product by a , a.c = α|a|2 +β(a.b) + γ [a a b]
⇒ |a|.|c|cosθ = α.1 + 0 + 0 ⇒ 1.|c|.cosθ = α
As |c| = 1 ; ∴ α = cosθ
Taking dot product of (i) by b
b . c = b . a + β | b|2 + γ [b a b] ⇒ | b||c|cos θ = 0 + β .1 + 0
∴ β = 1.1.cosθ = cosθ
|c|2 = 1 ⇒ α 2 + β 2 + γ 2 = 1 ⇒ cos2 θ + cos2 θ + γ 2 = 1
∴ γ 2 = 1 − 2cos2 θ = − cos 2θ
Hence, α = β = cosθ ,γ 2 = − cos 2θ
The locus of a point equidistant from two given points whose position vectors are a and b is equal to
(a) r − 1 (a + b) .(a + b) = 0 (b) r − 1 (a + b) .(a − b) = 0
2 2
(c) r − 1 (a + b) . a = 0 (d) [r − (a + b)]. b = 0
2
Let P (r) be a point on the locus.
∴ AP = BP P(r)
⇒ | r − a| =| r − b| ⇒ | r − a|2 =| r − b|2 ⇒ (r − a).(r − a) = (r − b).(r − b)
⇒ 2r .(a − b) = a .a − b.b ⇒ r .(a − b) = 1 (a + b).(a − b)
2
A(a) B(b)
1
∴ [r − 2 (a + b)].(a − b) = 0 . This is the locus of P.
Differential-equation
Many of the practical problems in physics and engineering can be converted into differential equations. The
solution of differential equations is, therefore of paramount importance. This chapter deals with some elementary
aspects of differential equations. These are addressed through simple application of differential and integral calculus.
There are two important aspects of differential equation, which have just been touched in this chapter. How to
formulate a problem as a differential equation is the one, and the other is how to solve it.
Definition .
An equation involving independent variable x, dependent variable y and the differential coefficients
dy , d2 y ,........ is called differential equation.
dx dx
2
Examples : (i) dy =1+ x+y (ii) dy + xy = cot x
dx dx
d4 y 3 dy d2y dy 2
dx dx dx2 + dx
(iii) 4 −4 + 4y = 5 cos 3x (iv) x2 1 + = 0
(1) Order of a differential equation : The order of a differential equation is the order of the highest
derivative occurring in the differential equation. For example, the order of above differential equations are 1,1,4
and 2 respectively.
Note : The order of a differential equation is a positive integer. To determine the order of a differential
equation, it is not needed to make the equation free from radicals.
(2) Degree of a differential equation : The degree of a differential equation is the degree of the highest
order derivative, when differential coefficients are made free from radicals and fractions. In other words, the degree
of a differential equation is the power of the highest order derivative occurring in differential equation when it is
written as a polynomial in differential coefficients.
Note : The definition of degree does not require variables x, y, t etc. to be free from radicals and
fractions. The degree of above differential equations are 1, 1, 3 and 2 respectively.
Example: 1 The order and degree of the differential equation y = x dy + a2 dy 2 + b2 are
Solution: (a) dx dx
Example: 2 (a) 1, 2 (b) 2, 1 (c) 1, 1 (d) 2, 2
Clearly, highest order derivative involved is dy , having order 1.
dx
Expressing the above differential equation as a polynomial in derivative, we have y − x dy 2 = a 2 dy 2 + b2
dx dx
(x2 − a2) dy 2 dy + y2 − b2
dx dx
i.e., − 2xy = 0
In this equation, the power of highest order derivative is 2. So its degree is 2.
d2y dy 1 1
dx 2 dx
The order and degree of the differential equation + 3 + x4 = 0 , are respectively
(a) 2, 3 (b) 3, 3 (c) 2, 6 (d) 2, 4
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