Question

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

Answer 1

Hint: For solving such a type of question we will assume $\cos x$ as the variable $'t'$. After that we will differentiate the equation and put that value into the equation given in the question. You will get a polynomial in $'t'$.then use partial fractions method for integrating.

Complete step-by-step answer:
Let $t = \cos x$
Differentiating both sides with respect to x, we will get,
$\therefore dt = - \sin xdx$
Putting value in question, we will get,
$ = - \int {\dfrac{{dt}}{{(1 - t)(2 - t)}}} $
Now, let us use partial fraction for further solving,
$\dfrac{1}{{(1 - t)(2 - t)}} = \dfrac{A}{{(1 - t)}} + \dfrac{B}{{(2 - t)}}.......(1)$
Taking LCM and adding in RHS, we will get,
$\therefore 1 = A(2 - t) + B(1 - t).........(2)$
Putting $t = 1$ in equation (2)
We will get $A = 1$
Putting $t = 2$ In equation (2)
We will get $B = - 1$
$A = 1$ or $B = - 1$
Putting above values in equation (1), we will get,
$\therefore \dfrac{1}{{(1 - t)(2 - t)}} = \dfrac{1}{{1 - t}} - \dfrac{1}{{2 - t}}$
Ingratiating both sides, we will get,
$\therefore \int {\dfrac{1}{{(1 - t)(2 - t)}}dt = \int {\dfrac{{dt}}{{1 - t}} + \int {\dfrac{{dt}}{{t - 2}}} } } $
$\therefore \int {\dfrac{{dt}}{{(1 - t)(2 - t)}} = - In(1 - t) + In(t - 2)} + c$
Now, putting $t = \cos x$, we will get,
$\therefore \int {\dfrac{{\sin x}}{{(1 - \cos x)(2 - \cos x)}}dx = In(\dfrac{{\cos x - 2}}{{1 - \cos x}})} + c$

Note: Don’t forget to add ‘c’(the constant of integration). In the question of indefinite integral, it is necessary to add a constant of integration in the answer. If you forget to add ‘c’ your answer will be wrong.