integration

392 INTEGRAL CALCULUS
Let u = (5x − 3) then du
dx = 5 and dx = du 5
Hence
∫
∫
4
4
(5x − 3) dx = du
u 5 = 4 5
∫ 1
u du
= 4 5 ln u + c = 4 ln(5x − 3) + c
5
Problem 4. Evaluate ∫ 1
0 2e6x−1 dx, correct to
4 significant figures.
Let u = 6x − 1 then du
dx = 6 and dx = du 6
Hence
∫ ∫
2e 6x−1 dx = 2e u du 6 = 1 ∫
e u du
3
Thus
∫ 1
0
= 1 3 eu + c = 1 3 e6x−1 + c
2e 6x−1 dx = 1 3 [e6x−1 ] 1 0 = 1 3 [e5 − e −1 ] = 49.35,
Problem 5.
correct to 4 significant figures.
Determine ∫ 3x(4x 2 + 3) 5 dx.
Let u = (4x 2 + 3) then du
du
= 8x and dx =
dx 8x
Hence
∫ ∫
3x(4x 2 + 3) 5 dx = 3x(u) 5 du
8x
= 3 ∫
u 5 du, by cancelling
8
The original variable ‘x’ has been completely
removed and the integral is now only in terms of
u and is a standard integral.
Hence 3 ∫
u 5 du = 3 ( u
6
)
+ c
8
8 6
= 1 16 u6 + c = 1
16 (4x2 + 3) 6 + c
Problem 6.
Evaluate
∫ π
6
0
24 sin 5 θ cos θ dθ.
Let u = sin θ then du = cos θ and dθ =
du
dθ
∫
∫
Hence 24 sin 5 θ cos θ dθ =
∫
= 24 u 5 du, by cancelling
Thus
∫ π
6
0
cos θ
24u 5 cos θ du
cos θ
= 24 u6
6 + c = 4u6 + c = 4(sin θ) 6 + c
= 4 sin 6 θ + c
24 sin 5 θ cos θ dθ = [4 sin 6 π
θ] 60
[ (
= 4 sin π ) 6
− ( sin 0) 6]
6
[ (1 ) 6
= 4 − 0]
= 1 or 0.0625
2
16
Now try the following exercise.
Exercise 154 Further problems on integration
using algebraic substitutions
In Problems 1 to 6, integrate with respect to the
variable.
1. 2 sin (4x + 9)
[− 1 ]
2 cos (4x + 9) + c
2. 3 cos (2θ − 5)
3. 4 sec 2 (3t + 1)
4.
5.
[ 3
2 sin (2θ − 5) + c ]
[ 4
3 tan (3t + 1) + c ]
[ ]
1
1
2 (5x − 3)6 70 (5x − 3)7 + c
−3
[− 3 ]
(2x − 1)
2 ln (2x − 1) + c
6. 3e 3θ+5 [e 3θ + 5 + c]
In Problems 7 to 10, evaluate the definite integrals
correct to 4 significant figures.