Put sinx = t
so, cosxdx = dt
∫

2dt

(1−t)(1+t2)

using partial fraction

2dt

(1−t)(1+t2)

=

A

(1−t)

+

Bt+C

(1+t2)

solving we get A = 1,B = 1,C = 1
∫

2dt

(1−t)(1+t2)

= ∫ [

1

(1−t)

+

t+1

(1+t2)

] dt
= ∫

1

(1−t)

dt + ∫

t

(1+t2)

dt + ∫

1

(1+t2)

dt
= - log (1 - t) +

1

2

log (1+t2) + tan−1 t + c
Re substituting ,
= log (

√1+sin2x

1−sinx

) + tan−1 (sin x) + c