Question

\[\int\frac{\sin x - \cos x}{\sqrt{\sin 2x}} dx\]

Answer 1

\[\int\left( \frac{\sin x - \cos x}{\sqrt{\sin 2x}} \right) dx\]
\[ = \int\left( \frac{\sin x - \cos x}{\sqrt{1 + \sin 2x - 1}} \right)dx\]
\[ = \int\frac{\left( \sin x - \cos x \right)}{\sqrt{\sin^2 x + \cos^2 x + 2 \sin x \cos x - 1}}dx\]
\[ = \int\frac{\left( \sin x - \cos x \right)}{\sqrt{\left( \sin x + \cos x \right)^2 - 1}}dx\]
\[\text{ let }\sin x + \cos x = t\]
\[ \Rightarrow \left( \cos x - \sin x \right) dx = dt\]
\[ \Rightarrow \left( \sin x - \cos x \right)dx = - dt\]
\[Now, \int\frac{\left( \sin x - \cos x \right)}{\sqrt{\left( \ sin x + \cos x \right)^2 - 1}}dx\]
\[ = - \int\frac{dt}{\sqrt{t^2 - 1^2}}\]
\[ = - \text{ log }\left| t + \sqrt{t^2 - 1} \right| + C\]
\[ = - \text{ log }\left| \sin x + \cos x + \sqrt{\left( \sin x + \cos x \right)^2 - 1} \right| + C\]
\[ = - \text{ log }\left| \sin x + \cos x + \sqrt{\sin^2 x + \cos^2 x + 2\sin  x . \cos x - 1} \right| + C\]
\[ = - \text{ log } \left| \sin x + \cos x + \sqrt{\sin 2x} \right| + C\]