Question

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

Answer 1

$$\begin{gathered} \int \left(\frac{\sin x+2\cos x}{2\sin x+\cos x}\right)dx \\ \mathrm{Let}\sin x+2\cos x=A\frac{d}{dx}\left(2\sin x+\cos x\right)+B\left(2\sin x+\cos x\right) \\ \Rightarrow \sin x+2\cos x=A\left(2\cos x-\sin x\right)+2\mathit{}B\sin \mathrm{x}+B\cos x \\ \Rightarrow \sin x+2\cos x=\left(2A+B\right)\cos x+\left(2B-A\right)\sin x \\ \mathrm{Equating}\mathrm{coefficients}\mathrm{of}\mathrm{like}\mathrm{terms} \\ \Rightarrow 2A+B=2...\left(1\right) \\ \Rightarrow -A+2B=1...\left(2\right) \\ \mathrm{Multiplying}\mathrm{eq}\left(2\right)\mathrm{by}2\mathrm{and}\mathrm{adding}\mathrm{it}\mathrm{to}\mathrm{eq}\left(1\right)\mathrm{we}\mathrm{get}, \\ 5B=4 \\ \Rightarrow B=\frac{4}{5} \\ \mathrm{Putting}B=\frac{4}{5}\mathrm{in}\mathrm{eq}\left(1\right)\mathrm{we}\mathrm{get}, \\ 2A+\frac{4}{5}=2 \\ \Rightarrow A=\frac{3}{5} \\ \therefore \int \left(\frac{\sin x+2\cos x}{2\sin x+\cos x}\right)dx=\int \left[\frac{{\displaystyle \frac{3}{5}}\left(2\cos x-\sin x\right)}{2\sin x+\cos x}\right]dx+\frac{4}{5}\int \frac{\left(2\sin x+\cos x\right)}{\left(2\sin x+\cos x\right)}dx \\ =\frac{3}{5}\int \left(\frac{2\cos x-\sin x}{2\sin x+\cos x}\right)dx+\frac{4}{5}\int dx \\ \mathrm{Putting}2\sin x+\cos x=t \\ \Rightarrow \left(2\cos x-\sin x\right)dx=dt \\ \therefore I=\frac{3}{5}\int \frac{dt}{t}+\frac{4}{5}\int dx \\ =\frac{3}{5}\ln \left|t\right|+\frac{4x}{5}+C \\ =\frac{3}{5}\ln \left|2\sin x+\cos x\right|+\frac{4x}{5}+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\because \mathit{}t=2\sin x\mathit{}+\cos x\right] \end{gathered}$$