Question

Write a value of\[\int\frac{\sin x - \cos x}{\sqrt{1 + \sin 2x}} \text{ dx}\]

Answer 1

\[\text{ Let I } = \int\frac{\left( \sin x + \cos x \right) dx}{\sqrt{1 - \sin 2x}}\]
\[ = \int\frac{\left( \sin x + \cos x \right) dx}{\sqrt{\sin^2 x + \cos^2 x - 2 \sin x \cos x}}\]
\[ = \int\frac{\left( \sin x + \cos x \right) dx}{\sqrt{\left( \sin x - \cos x \right)^2}}\]
\[ = \int\frac{\left( \sin x + \cos x \right) dx}{\left| \sin x - \cos x \right|}\]
\[ = \pm \int\left( \frac{\sin x + \cos x}{\sin x - \cos x} \right)dx\]
\[\text{ Let sin x} - \cos x = t\]
\[ \Rightarrow \left( \cos x + \sin x \right)dx = dt\]
\[ \therefore I = \pm \int\frac{dt}{t}\]
\[ = \pm \text{ ln }\left| t \right| + C\]
\[ = \pm \text{ ln} \left| \sin x - \cos x \right| + C \left( \because t = \sin x - \cos x \right)\]