394 Higher Engineering Mathematics
Now try the following exercise
Exercise 153 Further problems on
integration using algebraic substitutions
In Problems 1 to 6, integrate with respect to the
variable.
- 2sin( 4 x+ 9 )
[
−
1
2
cos( 4 x+ 9 )+c
]
- 3cos( 2 θ− 5 )
[
3
2
sin( 2 θ− 5 )+c
]
- 4sec^2 ( 3 t+ 1 )
[
4
3
tan( 3 t+ 1 )+c
]
4.
1
2
( 5 x− 3 )^6
[
1
70
( 5 x− 3 )^7 +c
]
5.
− 3
( 2 x− 1 )
[
−
3
2
ln( 2 x− 1 )+c
]
- 3e^3 θ+^5 [e^3 θ+^5 +c]
In Problems 7 to 10, evaluate the definite integrals
correct to 4 significant figures.
7.
∫ 1
0
( 3 x+ 1 )^5 dx [227.5]
8.
∫ 2
0
x
√
( 2 x^2 + 1 )dx [4.333]
9.
∫ π
3
0
2sin
(
3 t+
π
4
)
dt [0.9428]
10.
∫ 1
0
3cos( 4 x− 3 )dx [0.7369]
39.4 Further worked problems on
integration using algebraic
substitutions
Problem 7. Find
∫
x
2 + 3 x^2
dx.
Letu= 2 + 3 x^2 then
du
dx
= 6 xand dx=
du
6 x
Hence
∫
x
2 + 3 x^2
dx=
∫
x
u
du
6 x
=
1
6
∫
1
u
du,
by cancelling
=
1
6
lnu+c=
1
6
ln(2+ 3 x^2 )+c
Problem 8. Determine
∫
2 x
√
( 4 x^2 − 1 )
dx.
Letu= 4 x^2 −1then
du
dx
= 8 xand dx=
du
8 x
Hence
∫
2 x
√
( 4 x^2 − 1 )
dx=
∫
2 x
√
u
du
8 x
=
1
4
∫
1
√
u
du,by cancelling
=
1
4
∫
u
− 1
(^2) du=
1
4
⎡
⎢
⎣
u
(− 1
2
)
- 1
−
1
2
1
⎤
⎥
⎦+c
1
4
⎡
⎢
⎣
u
1
2
1
2
⎤
⎥
⎦+c=
1
2
√
u+c
1
2
√
(4x^2 −1)+c
Problem 9. Show that
∫
tanθdθ=ln(secθ)+c.
∫
tanθdθ=
∫
sinθ
cosθ
dθ.Letu=cosθ
then
du
dθ
=−sinθand dθ=
−du
sinθ
Hence
∫
sinθ
cosθ
dθ=
∫
sinθ
u
(
−du
sinθ
)
=−
∫
1
u
du=−lnu+c
=−ln(cosθ)+c=ln(cosθ)−^1 +c,
by the laws of logarithms.