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