Question

\[\int\frac{\cos x - \sin x}{1 + \sin 2x} dx\]

Answer 1

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