integration

394 INTEGRAL CALCULUS
It is possible in this case to change the limits of integration.
Thus when x = 3, u = 2(3) 2 + 7 = 25 and
when x = 1, u = 2(1) 2 + 7 = 9.
Hence
∫ x=3
x=1
5x √ (2x 2 + 7) dx =
= 5 4
∫ u=25
u=9
∫ 25
9
∫ 25
5x √ u du
4x
√ u du
= 5 u 2 1 du
4 9
Thus the limits have been changed, and it is unnecessary
to change the integral back in terms of x.
⎡
∫ x=3
Thus 5x √ (2x 2 + 7) dx = 5 ⎣ u 3 ⎤25
2
⎦
4 3/2
x=1
= 5 [√ 25
u 3]
6
= 5 [√
25
9 6
3 − √ ]
9 3
9
i.e. the limits have been changed
⎡ ⎤
= 3 ⎢
⎣ u 2
1 9
⎥
4 1
⎦ = 3 [√ √ ]
9 − 1 = 3,
2
2 1
taking positive values of square roots only.
Now try the following exercise.
Exercise 155 Further problems on integration
using algebraic substitutions
In Problems 1 to 7, integrate with respect to the
variable.
[ ]
1
1. 2x(2x 2 − 3) 5 12 (2x2 − 3) 6 + c
2. 5 cos 5 t sin t
[
− 5 ]
6 cos6 t + c
3. 3 sec 2 3x tan 3x
[ 1
2 sec2 3x + c or 1 2 tan2 3x + c]
Problem 11.
= 5 6 (125 − 27) = 812 3
Evaluate
∫ 2
0
3x
√
(2x 2 + 1) dx,
taking positive values of square roots only.
4. 2t √ (3t 2 − 1)
5.
ln θ
θ
6. 3 tan 2t
[ 2
9
√
]
(3t 2 − 1) 3 + c
[ 1
2 (lnθ)2 + c]
[ 3
2 ln ( sec 2t) + c ]
Let u = 2x 2 + 1 then du
dx
= 4x and dx =
du
4x
7.
2e t
√ (e t + 4)
[
4
√ (e t + 4) + c ]
Hence
∫ 2
0
∫
3x
x=2
√
(2x 2 + 1) dx = 3x du
√ u 4x
= 3 4
x=0
∫ x=2
x=0
u −1
2 du
In Problems 8 to 10, evaluate the definite integrals
correct to 4 significant figures.
8.
∫ 1
0
3x e (2x2 −1) dx [1.763]
Since u = 2x 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,
9.
10.
∫ π
2
0
∫ 1
0
3 sin 4 θ cos θ dθ [0.6000]
3x
(4x 2 dx [0.09259]
− 1) 5