394 Higher Engineering Mathematics
Now try the following exercise
Exercise 153 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
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 3θ+5
[e 3θ + 5 + c]
In Problems 7 to 10, evaluate the definite integrals
correct to 4 significant figures.
7.
1
0
(3x + 1)
5 dx
[227.5]
8.
2
0
x
(2x 2 + 1) dx
[4.333]
9.
π
3
0
2 sin
3t +
π
4
dt
[0.9428]
10.
1
0
3 cos(4x − 3) dx
[0.7369]
39.4 Further worked problems on
integration using algebraic
substitutions
Problem 7. Find
x
2 + 3x 2 dx.
Let u = 2 +3x 2 then
du
dx
= 6x and dx =
du
6x
Hence
x
2 + 3x 2 dx =
x
u
du
6x
=
1
6
1
u
du,
by cancelling
=
1
6
ln u + c =
1
6
ln(2 + 3x
2 ) + c
Problem 8. Determine
2x
(4x 2 − 1)
dx.
Let u = 4x
2
− 1 then
du
dx
= 8x and dx =
du
8x
Hence
2x
(4x 2 − 1)
dx =
2x
√
u
du
8x
=
1
4
1
√
u
du, by cancelling
=
1
4
u
−1
2 du =
1
4
⎡
⎢
⎣
u
−1
2
+1
−
1
2
+ 1
⎤
⎥
⎦ + c
=
1
4
⎡
⎢
⎣
u
1
2
1
2
⎤
⎥
⎦ + c =
1
2
√
u + c
=
1
2
(4x 2 − 1)+ c
Problem 9. Show that
tan θ dθ = ln(sec θ) + c.
tan θ dθ =
sin θ
cos θ
dθ. Let u = cos θ
then
du
dθ
= −sin θ and dθ =
−du
sin θ
Hence
sin θ
cos θ
dθ =
sin θ
u
−du
sin θ
= −
1
u
du = −ln u + c
= −ln(cos θ) + c = ln(cos θ)
−1
+ c,
by the laws of logarithms.