Integration using algebraicsubstitutions 393

be a lengthy process, and thus an algebraic substitution
is made.

Letu=( 2 x− 5 )then
du
dx

=2anddx=
du
2
Hence

∫

( 2 x− 5 )^7 dx=

∫

u^7

du

2

=

1

2

∫

u^7 du

=

1

2

(

u^8

8

)

+c=

1

16

u^8 +c

Rewritinguas( 2 x− 5 )gives:

∫

(2x−5)^7 dx=

1

16

(2x−5)^8 +c

Problem 3. Find

∫

4

( 5 x− 3 )

dx.

Letu=( 5 x− 3 )then

du

dx

=5anddx=

du
5
Hence

∫

4

( 5 x− 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−1then
du
dx
=6anddx=
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( 4 x^2 + 3 )^5 dx.
Letu=( 4 x^2 + 3 )then
du
dx
= 8 xand dx=
du
8 x
Hence
∫
3 x( 4 x^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 ofuand 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
24sin^5 θcosθdθ.
Letu=sinθthen
du
dθ
=cosθand dθ=
du
cosθ
Hence
∫
24sin^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
=4sin^6 θ+c
Thus
∫ π
6
0
24sin^5 θcosθdθ=[4sin^6 θ]
π
6
0
= 4
[(
sin
π
6
) 6
−(sin 0)^6
]
= 4
[(
1
2
) 6
− 0
]

1
16
or 0. 0625