Question:

$I = \int \frac{\sin x}{1 + \sin x \cos x} dx = \int \frac{\sin x + \cos x + \sin x - \cos x}{2 + \sin 2x} dx$<br>$= \int \frac{\sin x + \cos x}{3 - (\sin x - \cos x)^2} dx - \int \frac{\cos x - \sin x}{1 + (\cos x + \sin x)^2} dx$<br>$= \int \frac{dt}{3 - t^2} - \int \frac{du}{1 + u^2} \left[ \begin{matrix} t = \sin x - \cos x \\ u = \sin x + \cos x \end{matrix} \right] = \frac{1}{2\sqrt{3}} \ln \left| \frac{t + \sqrt{3}}{t - \sqrt{3}} \right| - \tan^{-1} u + C$<br>$= \frac{1}{2\sqrt{3}} \ln \left| \frac{\sin x - \cos x + \sqrt{3}}{\sin x - \cos x - \sqrt{3}} \right| - \tan^{-1} (\sin x + \cos x) + C$