Question

$\int \frac{\cos 2x}{{\left(\cos x+\sin x\right)}^2}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{\cos 2x}{{\left(\cos x+\sin x\right)}^2}dx \\ =\int \frac{{\cos }^{2}x-{\sin }^{2}x}{{\left(\cos x+\sin x\right)}^2}dx \\ =\int \frac{\cos x-\sin x}{\cos x+\sin x}dx \\ \mathrm{Putting}\cos x+\sin x=t \\ \Rightarrow -\sin x+\cos x=\frac{dt}{dx} \\ \Rightarrow \left(\cos x-\sin x\right)dx=dt \\ \therefore I=\int \frac{1}{t}dt \\ =\ln \left|t\right|+\mathrm{C} \\ =\ln \left|\cos x+\sin x\right|+\mathrm{C}\left[\because \mathit{}t=\cos x+\sin x\right] \end{gathered}$$