392 INTEGRAL CALCULUS

Letu=(5x−3) then

du

dx

=5 and dx=

du

5

Hence

∫

4

(5x−3)

dx=

∫

4

u

du

5

=

4

5

∫

1

u

du

=

4

5

lnu+c=

4

5

ln(5x−3)+c

Problem 4. Evaluate

∫ 1

0 2e

6 x− (^1) dx, correct to
4 significant figures.
Letu= 6 x−1 then
du
dx
=6 and dx=
du
6
Hence
∫
2e^6 x−^1 dx=
∫
2eu
du
6

1
3
∫
eudu

1
3
eu+c=
1
3
e^6 x−^1 +c
Thus
∫ 1
0
2e^6 x−^1 dx=
1
3
[e^6 x−^1 ]^10 =
1
3
[e^5 −e−^1 ]= 49. 35 ,
correct to 4 significant figures.
Problem 5. Determine
∫
3 x(4x^2 +3)^5 dx.
Letu=(4x^2 +3) then
du
dx
= 8 xand dx=
du
8 x
Hence
∫
3 x(4x^2 +3)^5 dx=
∫
3 x(u)^5
du
8 x

3
8
∫
u^5 du, by cancelling
The original variable ‘x’ has been completely
removed and the integral is now only in terms of
uand is a standard integral.
Hence
3
8
∫
u^5 du=
3
8
(
u^6
6
)
+c

1
16
u^6 +c=
1
16
(4x^2 +3)^6 +c
Problem 6. Evaluate
∫ π
6
0
24 sin^5 θcosθdθ.
Letu=sinθthen
du
dθ
=cosθand dθ=
du
cosθ
Hence
∫
24 sin^5 θcosθdθ=
∫
24 u^5 cosθ
du
cosθ
= 24
∫
u^5 du, by cancelling
= 24
u^6
6
+c= 4 u^6 +c=4(sinθ)^6 +c
=4 sin^6 θ+c
Thus
∫ π
6
0
24 sin^5 θcosθdθ=[4 sin^6 θ]
π
6
0
= 4
[(
sin
π
6
) 6
−( sin 0)^6
]
= 4
[(
1
2
) 6
− 0
]

1
16
or 0. 0625
Now try the following exercise.
Exercise 154 Further problems on integra-
tion 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

2

sin (2θ−5)+c

]

3. 4 sec^2 (3t+1)

[

4

3

tan (3t+1)+c

]

4.

1

2

(5x−3)^6

[

1

70

(5x−3)^7 +c

]

5.

− 3

(2x−1)

[

−

3

2

ln (2x−1)+c

]

6. 3e^3 θ+^5 [e^3 θ+^5 +c]

In Problems 7 to 10, evaluate the definite inte-

grals correct to 4 significant figures.