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.

1. 2x( 2 x^2 − 3 )^5

[

1

12

( 2 x^2 − 3 )^6 +c

]

2. 5cos^5 tsint

[

−

5

6

cos^6 t+c

]

3. 3sec^23 xtan3x
   [
   1
   2

sec^23 x+cor

1

2

tan^23 x+c

]

4. 2t

√

( 3 t^2 − 1 )

[

2

9

√

( 3 t^2 − 1 )^3 +c

]

5.

lnθ

θ

[

1

2

(lnθ)^2 +c

]