Question

Integrate the rational functions.
$\int \frac{cosx}{(1-sinx)(2-sinx)}dx.$

Answer 1

Let $I=\int \frac{cosx}{(1-sinx)(2-sinx)}dx$
Put $sinx=t\Rightarrow cosx=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{cosx}$
$\therefore I=\int \frac{cosx}{(1-t)(2-t)}\frac{dt}{cosx}=\int \frac{1}{(1-t)(2-t)}dt=\int [\frac{A}{1-t}+\frac{B}{2-t}]dt......\left(i\right)\therefore \frac{1}{(1-t)(2-t)}=\frac{A(2-t)+B(1-t)}{(1-t)(2-t)}\Rightarrow 1=2A-tA+B-Bt\Rightarrow 1=1(2A+B)+t(-A-B)$
On comparing the coefficients of t and constant term on both sides, we get
2A +B =1 and -A-B =0
On adding above equations, we get
A=1 and then B=1
$\therefore I=\int (\frac{1}{1-t}-\frac{1}{2-t})dt\text{[from Eq.(i)]}=\int \frac{1}{(1-t)}dt-\int \frac{1}{(2-t)}dt=\frac{log|1-t|}{(-1)}-\frac{log|2-t|}{(-1)}+C=log\left|\frac{2-t}{1-t}\right|+C=log\left|\frac{2-sinx}{1-sinx}\right|+C\text{(Put t =sin x)}$