Integration using algebraicsubstitutions 395
Hence
∫
tanθdθ=ln(secθ)+c,
since (cosθ)−^1 =
1
cosθ
=secθ
39.5 Change of limits
When evaluating definite integrals involving substi-
tutions it is sometimes more convenient tochange
the limitsof the integral as shown in Problems 10
and 11.
Problem 10. Evaluate
∫ 3
15 x
√
( 2 x^2 + 7 )dx,
taking positive values of square roots only.
Letu= 2 x^2 +7, then
du
dx
= 4 xand dx=
du
4 x
It is possible in this case to change the limits of inte-
gration. Thus whenx=3,u= 2 ( 3 )^2 + 7 =25 and when
x=1,u= 2 ( 1 )^2 + 7 =9.
Hence
∫x= 3
x= 1
5 x
√
( 2 x^2 + 7 )dx=
∫u= 25
u= 9
5 x
√
u
du
4 x
=
5
4
∫ 25
9
√
udu
=
5
4
∫ 25
9
u
1
(^2) du
Thus the limitshave been changed, and it is unnecessary
to change the integral back in terms ofx.
Thus
∫x= 3
x= 1
5 x
√
( 2 x^2 + 7 )dx=
5
4
⎡
⎣u
3
2
3 / 2
⎤
⎦
25
9
5
6
[√
u^3
] 25
9
5
6
[√
253 −
√
93
]
5
6
( 125 − 27 )= 81
2
3
Problem 11. Evaluate
∫ 2
0
3 x
√
( 2 x^2 + 1 )
dx,
taking positive values of square roots only.
Letu= 2 x^2 +1then
du
dx
= 4 xand dx=
du
4 x
Hence
∫ 2
0
3 x
√
( 2 x^2 + 1 )
dx=
∫x= 2
x= 0
3 x
√
u
du
4 x
3
4
∫x= 2
x= 0
u
− 1
(^2) du
Since u= 2 x^2 +1, when x= 2 ,u=9andwhen
x= 0 ,u=1.
Thus
3
4
∫x= 2
x= 0
u
− 1
(^2) du=^3
4
∫u= 9
u= 1
u
− 1
(^2) du,
i.e. the limits have been changed
3
4
⎡
⎢
⎣
u
1
2
1
2
⎤
⎥
⎦
9
1
3
2
[√
9 −
√
1
]
= 3 ,
taking positive values of square roots only.
Now try the following exercise
Exercise 154 Further problemson
integration using algebraic substitutions
In Problems 1 to 7, integrate with respect to the
variable.
- 2x( 2 x^2 − 3 )^5
[
1
12
( 2 x^2 − 3 )^6 +c
]
- 5cos^5 tsint
[
−
5
6
cos^6 t+c
]
- 3sec^23 xtan3x
[
1
2
sec^23 x+cor
1
2
tan^23 x+c
]
- 2t
√
( 3 t^2 − 1 )
[
2
9
√
( 3 t^2 − 1 )^3 +c
]
5.
lnθ
θ
[
1
2
(lnθ)^2 +c
]