The t = tan
θ
2 substitution 417
=
2 dt
1 + t 2
7(1 + t 2 ) − 3(2t ) + 6(1 − t 2 )
1 + t 2
=
2 dt
7 + 7t 2 − 6t + 6 − 6t 2
=
2 dt
t 2 − 6t + 13
=
2 dt
(t − 3) 2 + 2 2
= 2
1
2
tan
−1
t − 3
2
+ c
from 12, Table 40.1, page 399. Hence
dx
7 − 3 sin x + 6 cos x
= tan
−1
⎛
⎜
⎝
tan
x
2
− 3
2
⎞
⎟
⎠ + c
Problem 7. Determine
dθ
4 cosθ + 3 sin θ
From equations (1) to (3),
dθ
4 cos θ + 3 sinθ
=
2 dt
1 + t 2
4
1 − t 2
1 + t 2
+ 3
2t
1 + t 2
=
2 dt
4 − 4t 2 + 6t
=
dt
2 + 3t − 2t 2
= −
1
2
dt
t 2 −
3
2
t − 1
= −
1
2
dt
t −
3
4
2
−
25
16
=
1
2
dt
5
4
2
−
t −
3
4
2
=
1
2
⎡
⎢
⎢
⎣
1
2
5
4
ln
⎧
⎪ ⎪ ⎨
⎪ ⎪ ⎩
5
4
+
t −
3
4
5
4
−
t −
3
4
⎫
⎪ ⎪ ⎬
⎪ ⎪ ⎭
⎤
⎥
⎥
⎦ + c
from Problem 11, Chapter 41, page 413
=
1
5
ln
⎧
⎪ ⎨
⎪ ⎩
1
2
+ t
2 − t
⎫
⎪ ⎬
⎪ ⎭
+ c
Hence
dθ
4 cos θ + 3 sinθ
=
1
5
ln
⎧
⎪ ⎨
⎪ ⎩
1
2
+ tan
θ
2
2 − tan
θ
2
⎫
⎪ ⎬
⎪ ⎭
+ c
or
1
5
ln
⎧
⎪ ⎨
⎪ ⎩
1 +2 tan
θ
2
4 − 2 tan
θ
2
⎫
⎪ ⎬
⎪ ⎭
+ c
Now try the following exercise
Exercise 166 Further problems on the
t = tan
θ
2
substitution
In Problems 1 to 4, integrate with respect to the
variable.
1.
dθ
5 + 4 sinθ ⎡
⎢
⎣
2
3
tan −1
⎛
⎜
⎝
5 tan
θ
2
+ 4
3
⎞
⎟
⎠ + c
⎤
⎥
⎦
2.
d x
1 + 2 sin x
⎡
⎢
⎣
1
√
3
ln
⎧
⎪ ⎨
⎪ ⎩
tan
x
2
+ 2 −
√
3
tan
x
2
+ 2 +
√
3
⎫
⎪ ⎬
⎪ ⎭
+ c
⎤
⎥
⎦
3.
d p
3 − 4 sin p +2 cos p
⎡
⎢
⎣
1
√
11
ln
⎧
⎪ ⎨
⎪ ⎩
tan
p
2
− 4 −
√
11
tan
p
2
− 4 +
√
11
⎫
⎪ ⎬
⎪ ⎭
+ c
⎤
⎥
⎦