Integration using algebraicsubstitutions 395
Hence
tan θ dθ = ln(sec θ)+ c,
since
(cos θ) −1 =
1
cos θ
= sec θ
39.5 Change of limits
When evaluating definite integrals involving substitutions it is sometimes more convenient to change
the limits of the integral as shown in Problems 10
and 11.
Problem 10. Evaluate
3
1 5x
(2x 2 + 7) d x,
taking positive values of square roots only.
Let u =2x 2 + 7, then
du
dx
= 4x and dx =
du
4x
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 =
u=25
u=9
5x
√
u
du
4x
=
5
4
25
9
√
u du
=
5
4
25
9
u
1
2 du
Thus the limits have been changed, and it is unnecessary
to change the integral back in terms of x.
Thus
x=3
x=1
5x
(2x 2 + 7) dx =
5
4
⎡
⎣ u
3
2
3/2
⎤
⎦
25
9
=
5
6
u 3
25
9
=
5
6
√
25 3 −
√
9 3
=
5
6
(125 − 27) = 81
2
3
Problem 11. Evaluate
2
0
3x
(2x 2 + 1)
dx,
taking positive values of square roots only.
Let u =2x 2 + 1 then
du
dx
= 4x and dx =
du
4x
Hence
2
0
3x
(2x 2 + 1)
dx =
x=2
x=0
3x
√
u
du
4x
=
3
4
x=2
x=0
u
−1
2 du
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,
i.e. the limits have been changed
=
3
4
⎡
⎢
⎣
u
1
2
1
2
⎤
⎥
⎦
9
1
=
3
2
√
9 −
√
1
= 3,
taking positive values of square roots only.
Now try the following exercise
Exercise 154 Further problems on
integration using algebraic substitutions
In Problems 1 to 7, integrate with respect to the
variable.
1. 2x(2x
2
− 3)
5
1
12
(2x
2
− 3)
6
+ c
2. 5 cos 5 t sin t
−
5
6
cos 6 t + c
3. 3 sec
2 3x tan 3x
1
2
sec 2 3x + c or
1
2
tan 2 3x + c
4. 2t
(3t 2 − 1)
2
9
(3t 2 − 1) 3 + c
5.
ln θ
θ
1
2
(ln θ) 2 + c
Hence
tan θ dθ = ln(sec θ)+ c,
since
(cos θ) −1 =
1
cos θ
= sec θ
39.5 Change of limits
When evaluating definite integrals involving substitutions it is sometimes more convenient to change
the limits of the integral as shown in Problems 10
and 11.
Problem 10. Evaluate
3
1 5x
(2x 2 + 7) d x,
taking positive values of square roots only.
Let u =2x 2 + 7, then
du
dx
= 4x and dx =
du
4x
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 =
u=25
u=9
5x
√
u
du
4x
=
5
4
25
9
√
u du
=
5
4
25
9
u
1
2 du
Thus the limits have been changed, and it is unnecessary
to change the integral back in terms of x.
Thus
x=3
x=1
5x
(2x 2 + 7) dx =
5
4
⎡
⎣ u
3
2
3/2
⎤
⎦
25
9
=
5
6
u 3
25
9
=
5
6
√
25 3 −
√
9 3
=
5
6
(125 − 27) = 81
2
3
Problem 11. Evaluate
2
0
3x
(2x 2 + 1)
dx,
taking positive values of square roots only.
Let u =2x 2 + 1 then
du
dx
= 4x and dx =
du
4x
Hence
2
0
3x
(2x 2 + 1)
dx =
x=2
x=0
3x
√
u
du
4x
=
3
4
x=2
x=0
u
−1
2 du
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,
i.e. the limits have been changed
=
3
4
⎡
⎢
⎣
u
1
2
1
2
⎤
⎥
⎦
9
1
=
3
2
√
9 −
√
1
= 3,
taking positive values of square roots only.
Now try the following exercise
Exercise 154 Further problems on
integration using algebraic substitutions
In Problems 1 to 7, integrate with respect to the
variable.
1. 2x(2x
2
− 3)
5
1
12
(2x
2
− 3)
6
+ c
2. 5 cos 5 t sin t
−
5
6
cos 6 t + c
3. 3 sec
2 3x tan 3x
1
2
sec 2 3x + c or
1
2
tan 2 3x + c
4. 2t
(3t 2 − 1)
2
9
(3t 2 − 1) 3 + c
5.
ln θ
θ
1
2
(ln θ) 2 + c