Question

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

Answer 1

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