Question

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

Answer 1

$$\begin{gathered} \int \left(\frac{\cos x-\sin x}{1+\sin \left(2x\right)}\right)dx \\ \Rightarrow \int \left(\frac{\cos x-\sin x}{{\cos }^{2}x+{\sin }^{2}x+2\sin x.\cos x}\right)dx \\ \Rightarrow \int \frac{\left(\cos x-\sin x\right)}{{\left(\cos x+\sin x\right)}^2}dx \\ \mathrm{Let}\cos x+\sin x=t \\ \Rightarrow \left(-\sin x+\cos x\right)=\frac{dt}{dx} \\ \Rightarrow \left(-\sin x+\cos x\right)dx=dt \\ \mathrm{Now},\int \frac{\left(\cos x-\sin x\right)}{{\left(\cos x+\sin x\right)}^2}dx \\ =\int \frac{dt}{t^2} \\ =\int t^{-2}dt \\ =\frac{t^{-2+1}}{-2+1}+\mathrm{C} \\ =\frac{-1}{t}+\mathrm{C} \\ =-\frac{1}{\sin x+\cos x}+\mathrm{C} \end{gathered}$$