Question

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

Answer 1

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