Question

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

Answer 1

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

Let $t=\cos x$ $\therefore dt=-\sin xdx$

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

Now, $\frac{1}{(1-t)(2-t)}=\frac{A}{(1-t)}+\frac{B}{(2-t)}$

$\therefore 1=A(2-t)+B(1-t)$

$\therefore 2A+B=1$

$\therefore -A-B=0;A+B=0$

$A=1$ OR $B=-1$

$\therefore \frac{1}{(1-t)(2-t)}=\frac{1}{1-t}-\frac{1}{2-t}$

$\therefore \int \frac{1}{(1-t)(2-t)}dt=\int \frac{dt}{1-t}+\int \frac{dt}{t-2}$

$\therefore \int \frac{dt}{(1-t)(2-t)}=-ln(1-t)+ln(t-2)$

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