Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{\cos x}{{\left(1-\sin x\right)}^3\left(2+\sin x\right)}dx \\ \mathrm{Let},\sin x=t \\ \Rightarrow \cos xdx=dt \\ \mathrm{Now},\mathrm{integration}\mathrm{becomes}, \\ I=\int \frac{dt}{{\left(1-t\right)}^3\left(2+t\right)} \\ =-\int \frac{dt}{{\left(t-1\right)}^3\left(t+2\right)} \\ \mathrm{Let},\frac{1}{{\left(t-1\right)}^3\left(t+2\right)}=\frac{A}{\left(t-1\right)}+\frac{B}{{\left(t-1\right)}^2}+\frac{C}{{\left(t-1\right)}^3}+\frac{D}{\left(t+2\right)}.....\left(1\right) \\ \Rightarrow 1=A{\left(t-1\right)}^2\left(t+2\right)+B\left(t-1\right)\left(t+2\right)+C\left(t+2\right)+D{\left(t-1\right)}^3.....\left(2\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Putting}t=1\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ 1=3C \\ \Rightarrow C=\frac{1}{3} \\ \mathrm{Putting}t=-2\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ 1=D{\left(-2-1\right)}^3 \\ \Rightarrow 1=-27D \\ \Rightarrow D=\frac{-1}{27} \\ \mathrm{Putting}t=0\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ 1=2A-2B+2C-D \\ \Rightarrow 1=2A-2B+\frac{2}{3}+\frac{1}{27} \\ \Rightarrow 2A-2B=\frac{8}{27} \\ \Rightarrow A\mathit{-}B=\frac{4}{27} \\ \mathrm{Putting}t=2\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ 1=4A+4B+4C+D \\ \Rightarrow 1=4A+4B+\frac{4}{3}-\frac{1}{27} \\ \Rightarrow A+B=-\frac{2}{27} \\ \mathrm{Now},A\mathit{-}B=\frac{4}{27}\mathrm{and}A+B=-\frac{2}{27}\Rightarrow A=\frac{1}{27}\mathrm{and}B=\frac{-1}{9} \end{gathered}$$
$$\begin{gathered} \mathrm{Substituting}\mathrm{the}\mathrm{values}\mathrm{of}A\mathit{,}\mathit{}B\mathit{,}\mathit{}C\mathrm{and}D\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \frac{1}{{\left(t-1\right)}^3\left(t+2\right)}=\frac{1}{27\left(t-1\right)}-\frac{1}{9{\left(t-1\right)}^2}+\frac{1}{3{\left(t-1\right)}^3}+\frac{-1}{27\left(t+2\right)} \\ \mathrm{Now},\mathrm{integration}\mathrm{becomes} \\ \mathrm{I}=-\int \left[\frac{1}{27\left(t-1\right)}-\frac{1}{9{\left(t-1\right)}^2}+\frac{1}{3{\left(t-1\right)}^3}+\frac{-1}{27\left(t+2\right)}\right]dt \\ =-\left[\frac{1}{27}\log \left|t-1\right|+\frac{1}{9\left(t-1\right)}-\frac{1}{6{\left(t-1\right)}^2}-\frac{1}{27}\log \left|t+2\right|\right]+C \\ =-\frac{1}{27}\log \left|\sin x-1\right|-\frac{1}{9\left(\sin x-1\right)}+\frac{1}{6{\left(\sin x-1\right)}^2}+\frac{1}{27}\log \left|\sin x+2\right|+C \\ =-\frac{1}{27}\log \left|1-\sin x\right|+\frac{1}{9\left(1-\sin x\right)}+\frac{1}{6{\left(1-\sin x\right)}^2}+\frac{1}{27}\log \left|2+\sin x\right|+C \end{gathered}$$