394 INTEGRAL CALCULUS

It is possible in this case to change the limits of inte-

gration. Thus whenx=3,u=2(3)^2 + 7 =25 and

whenx=1,u=2(1)^2 + 7 =9.

Hence

∫x= 3

x= 1

5 x

√

(2x^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 limits have been changed, and it is unneces-
sary to change the integral back in terms ofx.
Thus
∫x= 3
x= 1
5 x
√
(2x^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
√
(2x^2 +1)
dx,
taking positive values of square roots only.
Letu= 2 x^2 +1 then
du
dx
= 4 xand dx=
du
4 x
Hence
∫ 2
0
3 x
√
(2x^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=9 and when
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 155 Further problems on integra-
tion using algebraic substitutions
In Problems 1 to 7, integrate with respect to the
variable.

1. 2x(2x^2 −3)^5

[

1

12

(2x^2 −3)^6 +c

]

2. 5 cos^5 tsint

[

−

5

6

cos^6 t+c

]

3. 3 sec^23 xtan 3x
   [
   1
   2

sec^23 x+cor

1

2

tan^23 x+c

]

4. 2t

√

(3t^2 −1)

[

2

9

√

(3t^2 −1)^3 +c

]

5.

lnθ

θ

[

1

2

(lnθ)^2 +c

]

6. 3 tan 2t

[

3

2

ln ( sec 2t)+c

]

7.

2et

√

(et+4)

[

4

√

(et+4)+c

]

In Problems 8 to 10, evaluate the definite inte-

grals correct to 4 significant figures.

8.

∫ 1

0

3 xe(2x

(^2) −1)
dx [1.763]
9.
∫ π
2
0
3 sin^4 θcosθdθ [0.6000]
10.
∫ 1
0
3 x
(4x^2 −1)^5
dx [0.09259]