2.4 integration of hyperbolic functions

Integration of Trigonometric Functions Integration of Hyperbolic Functions
. න = + . න = +
. න = − + . න = +
. න = + . න = +
. න = + . න = − +
. න = − + . න = − +
. න = − + . න = − +

MOOC MAT438/ UiTM

න 2 ℎ 4 ℎ 4

Solution :

න 2 ℎ 4 ℎ 4 = ℎ 4


= න 2 ℎ 4 ℎ 4 ∙ 4 ℎ 4 = 4 ℎ 4

12
= 2 න ℎ 4 = 4 ℎ 4

1 Integrate wrt u,
= 2 න using power rule
(because no more
1 2
= 2 2 + hyperbolic
function)

1 ℎ24 Replace = ℎ 4
4
= + #

MOOC MAT438/ UiTM

ℎ 1 ℎ 1

න
2
Solution :

ℎ 1 ℎ 1 = 1 = −1
න

2

= න ℎ ℎ ∙ − 2 = − −2 = − 1
2
2

= − 2

= − න ℎ ℎ

= − − ℎ + Integrate wrt u, Refer from 1
using hyperbolic rule appendix 13. න csch coth = − csch +
(because there is a
hyperbolic function) If = 1,
→ න csch coth = − csch +
= ℎ +

1
= ℎ + #

MOOC MAT438/ UiTM

3

න ℎ ℎ3

2

MOOC MAT438/ UiTM

3

න ℎ ℎ3

Solution : 2 If you are unable to memorize
the derivative formula, you can
3 retrieve from the integration
formula provided in the appendix
න ℎ ℎ3 = ℎ
(final examination paper)
2
1
3 12. න ℎ ℎ = − ℎ +
= − ℎ ℎ
= න ℎ2 ℎ ℎ If = 1, then
න ℎ ℎ = − ℎ +
2
= − ℎ ℎ Thus,
= 3
− ℎ ℎ ℎ = − ℎ ℎ
= න 2 ∙ ℎ ℎ ∙

= 2

=0.6 From = ℎ ,
When = 3, = ℎ 3 = 0.6
= − න 2 When = 2, = ℎ 2 = 0.8

=0.8

MOOC MAT438/ UiTM

Solution :

3 =0.6 Integrate wrt u, using
power rule (because no
න ℎ ℎ3 = − න 2
more hyperbolic
2 =0.8 function)

3 0.6
= − 3 0.8

0.6 3 0.8 3 0.8 3 0.6 3
there is no need to =− 3 − 3 = 3−3
replace = ℎ
because 0.6 and 0.8
are the values for u

= 0.0987 #

MOOC MAT438/ UiTM

‫׬‬ ℎ 2
3− ℎ 2

MOOC MAT438/ UiTM

ℎ 2
න 3 − ℎ 2

Solution :

ℎ 2 = 3 − ℎ 2
න 3 − ℎ 2

ℎ 2
= න ∙ −2 ℎ2 = −2 ℎ2

11
= − 2 න = −2 ℎ2

1 Integrate wrt u,
= − 2 + C using ln rule

(because no more
hyperbolic function)

1 Replace = 3 − ℎ 2
= − 2 3 − ℎ 2 + C #

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM