Question

$\int \frac{\sin 2x}{\left(1+\sin x\right)\left(2+\sin x\right)}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{\sin 2xdx}{\left(1+\sin x\right)\left(2+\sin x\right)} \\ =\int \frac{2\sin x\cos xdx}{\left(1+\sin x\right)\left(2+\sin x\right)} \\ \mathrm{Putting}\sin x=t \\ \Rightarrow \cos xdx=dt \\ \therefore I=\int \frac{2tdt}{\left(1+t\right)\left(2+t\right)} \\ =2\int \frac{tdt}{\left(1+t\right)\left(2+t\right)} \\ \mathrm{Let}\frac{t}{\left(1+t\right)\left(2+t\right)}=\frac{A}{1+t}+\frac{B}{2+t} \\ \Rightarrow \frac{t}{\left(1+t\right)\left(2+t\right)}=\frac{A\left(2+t\right)+B\left(1+t\right)}{\left(1+t\right)\left(2+t\right)} \\ \Rightarrow t=A\left(2+t\right)+B\left(1+t\right) \\ \mathrm{Putting}2+t=0 \\ \Rightarrow t=-2 \\ -2=A\times 0+B\left(-2+1\right) \\ \Rightarrow -2=B\left(-1\right) \\ \Rightarrow B=2 \\ \mathrm{let}t+1=0 \\ t=-1 \\ \Rightarrow -1=A\left(-1+2\right)+B\times 0 \\ A=-1 \\ \therefore I=2\int \left(\frac{-1}{t+1}+\frac{2}{t+2}\right)dt \\ =2\left[-\log \left|t+1\right|+2\log \left|t+2\right|\right]+C \\ =4\log \left|t+2\right|-2\log \left|t+1\right|+C \\ =\log \left|\frac{{\left(t+2\right)}^4}{{\left(t+1\right)}^2}\right|+C \\ =\log \left|\frac{{\left(\sin x+2\right)}^4}{{\left(\sin x+1\right)}^2}\right|+C \end{gathered}$$