integration

416 INTEGRAL CALCULUS
from 12, Table 40.1, page 398. Hence
∫
dx
7 − 3 sin x + 6 cos x
⎛
tan x ⎞
= tan −1 ⎜ 2 − 3 ⎟
⎝ ⎠ + c
2
Problem 7.
∫
Determine:
From equations (1) to (3),
∫
dθ
4 cos θ + 3 sin θ
2dt
∫
Hence
=
1 + t
( 2
1 − t
2
)
4
1 + t 2 + 3
∫
2dt
4 − 4t 2 + 6t =
dθ
4 cos θ + 3 sin θ
( 2t
1 + t 2 )
∫
dt
=
2 + 3t − 2t 2
=− 1 ∫
dt
2
t 2 − 3 2 t − 1
=− 1 ∫
dt
(
2
t − 3 ) 2
− 25
4 16
= 1 ∫
dt
( )
2 5 2 (
− t − 3 ) 2
4 4
⎡ ⎧
5
= 1 ⎢
1
⎪⎨
(t
4 + − 3 ) ⎫⎤
4
⎪⎬
2 ⎣
( ln 5 5
2
⎪⎩
(t
4)
4 − − 3 ) ⎥
⎦
⎪⎭
+ c
4
from problem 11, Chapter 41, page 411
⎧ ⎫
⎪⎨ 1
= 1 5 ln 2 + t ⎪⎬
⎪ ⎩ 2 − t ⎪⎭ + c
∫
dθ
4 cos θ + 3 sin θ
⎧
⎪⎨ 1
= 1 5 ln 2 + tan θ ⎫
⎪⎬
2
⎪ ⎩ 2 − tan θ ⎪⎭ + c
2
or
⎧
⎪⎨
1
1 + 2 tan θ ⎫
⎪⎬
5 ln 2
⎪ ⎩ 4 − 2 tan θ ⎪⎭ + c
2
Now try the following exercise.
Exercise 167 Further problems on the
t = tan θ/2 substitution
In Problems 1 to 4, integrate with respect to the
variable.
∫
dθ
1.
5 + 4 sin θ
⎡ ⎛
⎢2
5 tan θ ⎞ ⎤
⎜ ⎣
3 tan−1 2 + 4 ⎟ ⎥
⎝ ⎠ + c⎦
3
∫
dx
2.
1 + 2 sin x
⎡ ⎧
⎢
⎣√ 1 ⎪⎨ tan x
ln
2 + 2 − √ ⎫ ⎤
3⎪⎬
3 ⎪⎩ tan x 2 + 2 + √ 3
⎪⎭ + c ⎥
⎦
∫
dp
3.
3 − 4 sin p + 2 cos p
⎡ ⎧
⎢
⎣√ 1 ⎪⎨ tan p
ln
2 − 4 − √ ⎫ ⎤
11⎪⎬
11 ⎪⎩ tan p 2 − 4 + √ 11
⎪⎭ + c ⎥
⎦
∫
dθ
4.
3 − 4 sin θ
⎡ ⎧
⎢
⎣√ 1 ⎪⎨ 3 tan θ
ln
2 − 4 − √ ⎫ ⎤
7⎪⎬
7 ⎪⎩ 3 tan θ 2 − 4 + √ 7
⎪⎭ + c ⎥
⎦
5. Show that
⎧√ ⎫
∫
⎪⎨ t
dt
1 + 3 cos t = 1 2 + tan ⎪⎬
2 √ 2 ln 2
⎪√ ⎩
t 2 − tan
⎪⎭ + c
2
∫ π/3
3dθ
6. Show that
= 3.95, correct to 3
cos θ
0
significant figures.
7. Show that
∫ π/2
0
dθ
2 + cos θ = π
3 √ 3