Integration using algebraicsubstitutions 393
be a lengthy process, and thus an algebraic substitution
is made.
Let u =(2x − 5) then
du
dx
= 2 and dx =
du
2
Hence
(2x − 5)
7 dx =
u
7 du
2
=
1
2
u
7 du
=
1
2
u 8
8
+ c =
1
16
u
8
+ c
Rewriting u as (2x − 5) gives:
(2x − 5)
7 dx =
1
16
(2x −5)
8
+ c
Problem 3. Find
4
(5x − 3)
dx.
Let u =(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
ln u + c =
4
5
ln(5x −3)+ c
Problem 4. Evaluate
1
0 2e 6x−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
3
e
u du
=
1
3
e
u
+ c =
1
3
e
6x−1
+ c
Thus
1
0
2e
6x−1 dx =
1
3
[e
6x−1 ]
1
0 =
1
3
[e
5
− e
−1 ] = 49.35,
correct to 4 significant figures.
Problem 5. Determine
3x(4x 2 + 3) 5 dx.
Let u =(4x 2 + 3) then
du
dx
= 8x and dx =
du
8x
Hence
3x(4x
2
+ 3)
5 dx =
3x(u)
5 du
8x
=
3
8
u
5 du, by cancelling
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
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
θ cos θ dθ.
Let u = sin θ then
du
dθ
= cos θ and dθ =
du
cos θ
Hence
24 sin
5
θ cos θ dθ =
24u
5 cos θ
du
cos θ
= 24
u
5 du, by cancelling
= 24
u 6
6
+ c = 4u 6 + c = 4(sin θ) 6 + c
= 4 sin 6 θ + c
Thus
π
6
0
24 sin
5
θ cos θ dθ = [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