Thet=tanθ 2 substitution 417
=
∫
2dt
1 +t^2
7 ( 1 +t^2 )− 3 ( 2 t)+ 6 ( 1 −t^2 )
1 +t^2
=
∫
2dt
7 + 7 t^2 − 6 t+ 6 − 6 t^2
=
∫
2dt
t^2 − 6 t+ 13
=
∫
2dt
(t− 3 )^2 + 22
= 2
[
1
2
tan−^1
(
t− 3
2
)]
+c
from 12, Table 40.1, page 399. Hence
∫
dx
7 −3sinx+6cosx
=tan−^1
⎛
⎜
⎝
tan
x
2
− 3
2
⎞
⎟
⎠+c
Problem 7. Determine
∫
dθ
4cosθ+3sinθ
From equations (1) to (3),
∫
dθ
4cosθ+3sinθ
=
∫ 2dt
1 +t^2
4
(
1 −t^2
1 +t^2
)
+ 3
(
2 t
1 +t^2
)
=
∫
2dt
4 − 4 t^2 + 6 t
=
∫
dt
2 + 3 t− 2 t^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
4cosθ+3sinθ
=
1
5
ln
⎧
⎪⎨
⎪⎩
1
2
+tan
θ
2
2 −tan
θ
2
⎫
⎪⎬
⎪⎭
+c
or
1
5
ln
⎧
⎪⎨
⎪⎩
1 +2tan
θ
2
4 −2tan
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 +4sinθ
⎡
⎢
⎣
2
3
tan−^1
⎛
⎜
⎝
5tan
θ
2
+ 4
3
⎞
⎟
⎠+c
⎤
⎥
⎦
2.
∫
dx
1 +2sinx
⎡
⎢
⎣
1
√
3
ln
⎧
⎪⎨
⎪⎩
tan
x
2
+ 2 −
√
3
tan
x
2
+ 2 +
√
3
⎫
⎪⎬
⎪⎭
+c
⎤
⎥
⎦
3.
∫
dp
3 −4sinp+2cosp
⎡
⎢
⎣
1
√
11
ln
⎧
⎪⎨
⎪⎩
tan
p
2
− 4 −
√
11
tan
p
2
− 4 +
√
11
⎫
⎪⎬
⎪⎭
+c
⎤
⎥
⎦