Chapter 40
Integration using
trigonometric and
hyperbolic substitutions
40.1 Introduction
Table 40.1 gives a summary of the integrals that require
the use of trigonometric and hyperbolic substitutions
and their application is demonstrated in Problems 1
to 27.
40.2 Worked problems on integration
of sin
2 x, cos 2 x, tan 2 x and cot 2 x
Problem 1. Evaluate
π
4
0
2 cos
2 4t dt.
Since cos 2t = 2 cos 2 t − 1 (from Chapter 17),
then cos 2 t =
1
2
(1 + cos 2t ) and
cos 2 4t =
1
2
(1 + cos 8t )
Hence
π
4
0
2 cos
2 4t dt
= 2
π
4
0
1
2
(1 + cos 8t ) dt
=
t +
sin 8t
8
π
4
0
=
⎡
⎢
⎣
π
4
+
sin 8
π
4
8
⎤
⎥
⎦ −
0 +
sin 0
8
=
π
4
or 0.7854
Problem 2. Determine
sin 2 3x dx.
Since cos 2x = 1 − 2 sin 2 x (from Chapter 17),
then sin 2 x =
1
2
(1 − cos 2x) and
sin 2 3x =
1
2
(1 − cos 6x)
Hence
sin
2 3x dx =
1
2
(1 − cos 6x) dx
=
1
2
x −
sin 6x
6
+ c
Problem 3. Find 3
tan
2 4x dx.
Since 1 + tan 2 x = sec 2 x, then tan 2 x = sec 2 x − 1 and
tan 2 4x = sec 2 4x − 1.
Hence 3
tan
2 4x dx = 3
(sec
2 4x − 1) dx
= 3
tan 4x
4
− x
+ c