Thet=tanθ 2 substitution 415
42.2 Worked problemson the
t=tan
θ
2
substitution
Problem 1. Determine
∫
dθ
sinθ
Ift=tan
θ
2
then sinθ=
2 t
1 +t^2
anddθ=
2dt
1 +t^2
from
equations (1) and (3).
Thus
∫
dθ
sinθ
=
∫
1
sinθ
dθ
=
∫^1
2 t
1 +t^2
(
2 dt
1 +t^2
)
=
∫
1
t
dt=lnt+c
Hence
∫
dθ
sinθ
=ln
(
tan
θ
2
)
+c
Problem 2. Determine
∫
dx
cosx
If tan
x
2
then cosx=
1 −t^2
1 +t^2
and dx=
2dt
1 +t^2
from
equations (2) and (3).
Thus
∫
dx
cosx
=
∫
1
1 −t^2
1 +t^2
(
2dt
1 +t^2
)
=
∫
2
1 −t^2
dt
2
1 −t^2
may be resolved into partial fractions (see
Chapter 2).
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 = 2 A, from which,A= 1
When t=− 1 , 2 = 2 B, from which,B= 1
Hence
∫
2dt
1 −t^2
=
∫
1
( 1 −t)
+
1
( 1 +t)
dt
=−ln( 1 −t)+ln( 1 +t)+c
=ln
{
( 1 +t)
( 1 −t)
}
+c
Thus
∫
dx
cosx
=ln
⎧
⎪⎨
⎪⎩
1 +tan
x
2
1 −tan
x
2
⎫
⎪⎬
⎪⎭
+c
Note that since tan
π
4
=1, the above result may be
written as:
∫
dx
cosx
=ln
⎧
⎪⎨
⎪⎩
tan
π
4
+tan
x
2
1 −tan
π
4
tan
x
2
⎫
⎪⎬
⎪⎭
+c
=ln
{
tan
(π
4
+
x
2
)}
+c
from compound angles, Chapter 17.
Problem 3. Determine
∫
dx
1 +cosx
If tan
x
2
then cos x=
1 −t^2
1 +t^2
and dx=
2dt
1 +t^2
from
equations (2) and (3).
Thus
∫
dx
1 +cosx
=
∫
1
1 +cosx
dx
=
∫
1
1 +
1 −t^2
1 +t^2
(
2dt
1 +t^2
)
=
∫
1
( 1 +t^2 )+( 1 −t^2 )
1 +t^2
(
2dt
1 +t^2
)
=
∫
dt
Hence
∫
dx
1 +cosx
=t+c=tan
x
2
+c
Problem 4. Determine
∫
dθ
5 +4cosθ