Integrals — Class 12 Maths Solution

: Let $I = \int {\cfrac{{\cos xdx}}{{\left( {1 - \sin x} \right)\left( {2 - \sin x} \right)}}}$

Put $\sin x = t$ $\Rightarrow$ $\cos x\,dx = dt$

$\therefore$ $\int {\cfrac{{\cos \,x}}{{\left( {1 - \sin x} \right)\left( {2 - \sin x} \right)}}dx} = \int {\cfrac{1}{{\left( {1 - t} \right)\left( {2 - t} \right)}}dt}$

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

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

…(i)
Putting $t = 1$ in (i),

we get $1 = A\left( {2 - 1} \right)$ $\Rightarrow$ $A = 1$
Putting $t = 2$ in (i),

we get $B = - 1$
$\therefore$ $I = \int {\cfrac{{\cos xdx}}{{\left( {1 - \sin x} \right)\left( {2 - \sin x} \right)}}}$

$= \int {\cfrac{1}{{1 - t}}dt} - \int {\cfrac{{dt}}{{2 - t}}} = - \log \left| {1 - t} \right| + \log \left| {2 - t} \right| + C$

$= \log \left| {\cfrac{{2 - t}}{{1 - t}}} \right| + C = \log \left| {\cfrac{{2 - \sin \,x}}{{1 - \sin x}}} \right| + C$