UPSC NDA Math Model Paper 6

Here we have to find the value of ∫

sin‌x

1−4cos2x

dx
Let cos‌x=t then −sin‌x‌d‌x=dt
⇒∫

sin‌x

1−4cos2x

dx=−∫

dt

1−4t2

The integrand ∫

dt

1−4t2

can be written as:

⇒∫

dt

1−4t2

=

1

4

‌∫

dt

( 1 4 −t2)

=

1

4

.∫

dt

{( 1 2 )2−t2}

So, by comparing ∫

dt

{( 1 2 )2−t2}

with ∫

dx

a2−x2

we get: a=1∕2
As we know that, ∫

dx

a2−x2

=

1

2a

‌log|

a+x

a−x

|+C where C is a constant
⇒∫

dt

{( 1 2 )2−t2}

=

1

2×( 1 2 )

.log|

1 2 +t

1 2 −t

|
⇒−∫

dx

1−4t2

=−

1

4

‌log|

1+2t

1−2t

|+C
By substituting cos x = t in the above equation we get
⇒∫

sin‌x

1−4cos2x

dx =−

1

4

‌log|

1+2‌cos‌x

1−2‌cos‌x

|+C