Question

Evaluate $\int \frac{\cos xdx}{(1-\sin x)(2-\sin x)}$

Answer 1

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

Putting $sinx=t$

$\Rightarrow \cos xdx=dt$

$=\int \frac{dt}{\left(1-t\right)\left(2-t\right)}$

$=\int \frac{dt}{\left(t-1\right)\left(t-2\right)}$

$\frac{1}{\left(t-1\right)\left(t-2\right)}=\frac{A}{t-1}+\frac{B}{t-2}$

$\Rightarrow 1=A\left(t-2\right)+B\left(t-2\right)$

$\Rightarrow 1=At-2A+Bt-B$

$\Rightarrow 1=\left(A+B\right)t-2A-B$

On Comparing both sides

$A+B=0$ $-2A-B=1$

$\to A=-B$ $\Rightarrow 2A-B=1$

$\Rightarrow B=1$

So $A=-1$

$B=1$

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

$=-\log \left|t-1\right|+\log \left|t-2\right|+c$

$=\log \left|\sin x-2\right|-\log \left|\sin x-1\right|+c$

$=\log \left|\frac{\sin x-2}{\sin x-1}\right|+c$

$=\log \left|\frac{2-\sin x}{1-\sin x}\right|+c$