Put sinx = t so, cosxdx = dt ∫ $\frac{2dt}{(1-t)(1+t^2)}$ using partial fraction $\frac{2dt}{(1-t)(1+t^2)}$ = $\frac{A}{1-t}$ + $\frac{Bt+C}{1+t^2}$ solving we get A = 1,B = 1,C = 1 ∫ $\frac{2dt}{(1-t)(1+t^2)}$ = ∫ $\left[ \frac{1}{1-t} + \frac{t+1}{1+t^2} \right]$ dt = ∫ $\frac{1}{1-t}$ dt + ∫ $\frac{t}{1+t^2}$ dt + ∫ $\frac{1}{1+t^2}$ dt = - log (1 - t) + $\frac{1}{2}$ log $(1+t^2)$ + $\tan^{-1}$ t + c Re substituting , = log $\left( \frac{\sqrt{1+\sin^2 x}}{1-\sin x} \right)$ + $\tan^{-1}$ (sin x) + c