Question

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

 

 

Answer 1

\[\text{ Let I } = \int\left( \frac{\sin x - \cos x}{\sqrt{\sin 2x}} \right) dx\]
\[\text{ Putting sin x +  cos x = t}\]
\[ \Rightarrow \left( \cos x - \sin x \right) dx = dt\]
\[ \Rightarrow \left( \sin x - \cos x \right) dx = - dt\]
\[\text{ Also  sin x +  cos x = t}\]
\[\text{ Squaring both sides,} \]
\[ \left( \sin x + \cos x \right)^2 = t^2 \]
\[ \Rightarrow \sin^2 x + \cos^2 x + 2 \sin x \cos x = t^2 \]
\[ \Rightarrow 1 + \text{ sin  2x }= t^2 \]
\[ \Rightarrow \text{  sin  2x} = t^2 - 1\]
\[ \therefore I = \int\frac{- dt}{\sqrt{t^2 - 1}}\]
\[ = - \text{ ln} \left| t + \sqrt{t^2 - 1} \right| + C ..........\left( \because \int\frac{dt}{\sqrt{x^2 - a^2}} = \text{ ln}\left| x + \sqrt{x^2 - a^2} \right| + C \right)\]
\[ = - \text{ ln} \left| \left( \sin x + \cos x \right) + \sqrt{\left( \sin x + \cos x \right)^2 - 1} \right| + C ..........\left( \because t = \sin x + \cos x \right)\]
\[ = - \text{ ln }\left| \left( \sin x + \cos x \right) + \sqrt{\sin^2 x + \cos^2 x + 2 \sin \cos x - 1} \right| + C\]
\[ = - \text{ ln }\left| \sin x + \cos x + \sqrt{\sin 2 x} \right| + C\]