Question

$\int \frac{1}{\sin x\left(2+3\cos x\right)}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{1}{\sin x\left(2+3\cos x\right)}dx \\ =\int \frac{\sin x}{{\sin }^{2}x\left(2+3\cos x\right)}dx \\ =\int \frac{\sin x}{\left(1-{\cos }^{2}x\right)\left(2+3\cos x\right)}dx \\ =\int \frac{\sin x}{\left(1-\cos x\right)\left(1+\cos x\right)\left(2+3\cos x\right)}dx \\ \mathrm{Putting}\cos x=t \\ \Rightarrow -\sin xdx=dt \\ \therefore I=\int \frac{-1}{\left(1-t\right)\left(1+t\right)\left(2+3t\right)}dt \\ =\int \frac{1}{\left(t-1\right)\left(t+1\right)\left(3t+2\right)}dt \\ \mathrm{Let}\frac{1}{\left(t-1\right)\left(t+1\right)\left(3t+2\right)}=\frac{A}{t-1}+\frac{B}{t+1}+\frac{C}{3t+2} \\ \Rightarrow \frac{1}{\left(t-1\right)\left(t+1\right)\left(3t+2\right)}=\frac{A\left(t+1\right)\left(3t+2\right)+B\left(t-1\right)\left(3t+2\right)+C\left(t+1\right)\left(\mathrm{t}-1\right)}{\left(t-1\right)\left(t+1\right)\left(3t+2\right)} \\ \Rightarrow 1=A\left(t+1\right)\left(3t+2\right)+B\left(t-1\right)\left(3t+2\right)+C\left(t+1\right)\left(t-1\right) \\ \mathrm{Putting}t+1=0\mathrm{or}t=-1 \\ \Rightarrow 1=A\times 0+B\mathit{}\left(-1-1\right)\left(3\times -1+2\right)+C\times 0 \\ \therefore B=\frac{1}{2} \\ \mathrm{Now},\mathrm{putting}t-1=0\mathrm{or}t=1 \\ \Rightarrow 1=A\mathit{}\left(1+1\right)\left(3+2\right)+B\times 0+C\times 0 \\ \therefore A=\frac{1}{10} \\ \mathrm{Now},\mathrm{putting}3t+2=0\mathrm{or}t=\frac{-2}{3} \\ \Rightarrow 1=A\times 0+B\times 0+C\left(-\frac{2}{3}+1\right)\left(-\frac{2}{3}-1\right) \\ \Rightarrow 1=C\left(\frac{1}{3}\right)\left(\frac{-5}{3}\right) \\ \therefore C=\frac{-9}{5} \\ \therefore I=\int \frac{1}{10\left(t-1\right)}dt+\frac{1}{2}\int \frac{1}{t+1}dt-\frac{9}{5}\int \frac{1}{3t+2}dt \\ =\frac{1}{10}\ln \left|t-1\right|+\frac{1}{2}\ln \left|t+1\right|-\frac{9}{5}\ln \frac{\left|3t+2\right|}{3}+C \\ =\frac{1}{10}\ln \left|t-1\right|+\frac{1}{2}\log \left|t+1\right|-\frac{3}{5}\ln \left|3t+2\right|+C \\ =\frac{1}{10}+\ln \left|\cos x-1\right|+\frac{1}{2}\ln \left|\cos x+1\right|-\frac{3}{5}\ln \left|3\cos x+2\right|+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\because \mathit{}t=\cos \mathit{}x\right] \end{gathered}$$