Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{\cos x-\sin x}{1+\sin 2x}dx \\ =\int \frac{\cos x-\sin x}{{\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x}dx \\ =\int \frac{\cos x-\sin x}{{\left(\sin x+\cos x\right)}^2}dx \\ \mathrm{Putting}\sin x+\cos x=t \\ \Rightarrow \left(\cos x-\sin x\right)dx=dt \\ \therefore I=\int \frac{1}{t^2}dt \\ =-\frac{1}{t}+C \\ =-\frac{1}{\sin x+\cos x}+C \end{gathered}$$