416 Higher Engineering Mathematics
If t = tan
θ
2
then cos θ =
1 −t 2
1 +t 2 and dx =
2 dt
1 + t 2
from equations (2) and (3).
Thus
dθ
5 + 4 cosθ
=
2 dt
1 + t 2
5 + 4
1 − t 2
1 + t 2
=
2 dt
1 + t 2
5(1 + t 2 ) + 4(1 − t 2 )
(1 + t 2 )
= 2
dt
t 2 + 9
= 2
dt
t 2 + 3 2
= 2
1
3
tan
−1 t
3
+ c,
from 12 of Table 40.1, page 399. Hence
dθ
5 +4 cos θ
=
2
3
tan
−1
1
3
tan
θ
2
+ c
Now try the following exercise
Exercise 165 Further problems on the
t =tan
θ
2
substitution
Integrate the following with respect to the variable:
1.
dθ
1 + sin θ
⎡
⎢
⎣
−2
1 + tan
θ
2
+ c
⎤
⎥
⎦
2.
dx
1 − cos x + sin x ⎡
⎢
⎣ln
⎧
⎪ ⎨
⎪ ⎩
tan
x
2
1 + tan
x
2
⎫
⎪ ⎬
⎪ ⎭
+ c
⎤
⎥
⎦
3.
dα
3 + 2 cosα 2
√
5
tan
−1
1
√
5
tan
α
2
+ c
4.
dx
3 sin x − 4 cos x
⎡
⎢
⎣
1
5
ln
⎧
⎪ ⎨
⎪ ⎩
2 tan
x
2
− 1
tan
x
2
+ 2
⎫
⎪ ⎬
⎪ ⎭
+ c
⎤
⎥
⎦
42.3 Further worked problems on the
t = tan
θ
2
substitution
Problem 5. Determine
dx
sin x + cos x
If tan
x
2
then sin x =
2t
1 + t 2 , cosx =
1 − t 2
1 + t 2 and
d x =
2 dt
1 + t 2 from equations (1), (2) and (3).
Thus
dx
sin x + cos x
=
2 dt
1 + t 2
2t
1 + t 2
+
1 − t 2
1 + t 2
=
2 dt
1 + t 2
2t + 1 − t 2
1 + t 2
=
2 dt
1 + 2t − t 2
=
−2 dt
t 2 − 2t − 1
=
−2 dt
(t − 1) 2 − 2
=
2 dt
(
√
2) 2 − (t − 1) 2
= 2
1
2
√
2
ln
√
2 + (t − 1)
√
2 − (t − 1)
+ c
(see Problem 11, Chapter 41, page 413),
i.e.
dx
sin x + cos x
=
1
√
2
ln
⎧
⎪ ⎨
⎪ ⎩
√
2 − 1 +tan
x
2
√
2 + 1 −tan
x
2
⎫
⎪ ⎬
⎪ ⎭
+ c
Problem 6. Determine
dx
7 − 3 sin x + 6 cos x
From equations (1) and (3),
dx
7 − 3 sin x + 6 cos x
=
2 dt
1 + t 2
7 − 3
2t
1 + t 2
+ 6
1 − t 2
1 + t 2