integration

414 INTEGRAL CALCULUS
Hence
∫
Problem 2.
∫
1
( )
=
2t 2 dt
1 + t 2 1 + t
∫ 2 1
= dt = ln t + c
t
dθ
sin θ = ln (tan θ 2
∫
Determine:
)
+ c
dx
cos x
If tan x 1 − t2 2dt
then cos x = and dx =
2 1 + t2 1 + t 2 from
equations (2) and (3).
Thus
∫
∫
dx
cos x =
∫
=
1
( )
1 − t 2 2dt
1 + t 2 1 + t 2
2
1 − t 2 dt
2
may be resolved into partial fractions (see
1 − t2 Chapter 3).
Let
2
1 − t 2 = 2
(1 − t)(1 + t)
= A
(1 − t) + B
(1 + t)
A(1 + t) + B(1 − t)
=
(1 − t)(1 + t)
Hence 2 = A(1 + t) + B(1 − t)
When t = 1, 2 = 2A, from which, A = 1
When t =−1, 2 = 2B, from which, B = 1
∫ ∫
2dt
Hence
1 − t 2 = 1
(1 − t) + 1
(1 + t) dt
Thus
∫
=−ln(1 − t) + ln(1 + t) + c
{ } (1 + t)
= ln + c
(1 − t)
⎧
⎪⎨
dx
1 + tan x ⎫
⎪⎬
cos x = ln 2
⎪ ⎩ 1 − tan x ⎪⎭ + c
2
Note that since tan π = 1, the above result may be
4
written as:
∫
⎧
⎪⎨
dx
tan π
cos x = ln 4 + tan x 2
⎪ ⎩ 1 − tan π 4 tan x 2
{
= ln tan
( π
4 + x 2)}
+ c
from compound angles, Chapter 18.
Problem 3.
∫
Determine:
dx
1 + cos x
⎫
⎪⎬
⎪⎭ + c
If tan x 1 − t2 2dt
then cos x = and dx =
2 1 + t2 1 + t 2 from
equations (2) and (3).
∫
∫
dx
Thus
1 + cos x = 1
1 + cos x dx
∫
( )
1 2dt
=
1 + 1 − t2 1 + t 2
∫
1 + t 2 ( )
1
2dt
=
(1 + t 2 ) + (1 − t 2 ) 1 + t 2
∫ 1 + t 2
= dt
Hence
∫
dx
1 + cos x = t + c = tan x 2 + c
Problem 4.
∫
Determine:
dθ
5 + 4 cos θ
If t = tan θ 1 − t2
2dt
then cos θ = and dx =
2 1 + t2 1 + t 2
from equations (2) and (3).
( ) 2dt
∫
∫
dθ
Thus
5 + 4 cos θ = 1 + t 2
( 1 − t
2
)
5 + 4
1 + t
( 2 ) 2dt
∫
1 + t
=
2 5(1 + t 2 ) + 4(1 − t 2 )
∫ (1 + t 2 ∫)
dt
= 2
t 2 + 9 = 2 dt
(
t 2 + 3 2
1
= 2
3 3)
t tan−1 + c,