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Question: Evaluate \(\int {\dfrac{{\sin x}}{{(1 - \cos x)(2 - \cos x)}}} dx\)...

Evaluate ∫sin⁡x(1−cos⁡x)(2−cos⁡x)dx\int {\dfrac{{\sin x}}{{(1 - \cos x)(2 - \cos x)}}} dx

Explanation

Solution

For solving such a type of question we will assume cos⁡x\cos x as the variable ′t′'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′'t'.then use partial fractions method for integrating.

Complete step-by-step answer:
Let t=cos⁡xt = \cos x
Differentiating both sides with respect to x, we will get,
∴dt=−sin⁡xdx\therefore dt = - \sin xdx
Putting value in question, we will get,
=−∫dt(1−t)(2−t)= - \int {\dfrac{{dt}}{{(1 - t)(2 - t)}}}
Now, let us use partial fraction for further solving,
1(1−t)(2−t)=A(1−t)+B(2−t).......(1)\dfrac{1}{{(1 - t)(2 - t)}} = \dfrac{A}{{(1 - t)}} + \dfrac{B}{{(2 - t)}}.......(1)
Taking LCM and adding in RHS, we will get,
∴1=A(2−t)+B(1−t).........(2)\therefore 1 = A(2 - t) + B(1 - t).........(2)
Putting t=1t = 1 in equation (2)
We will get A=1A = 1
Putting t=2t = 2 In equation (2)
We will get B=−1B = - 1
A=1A = 1 or B=−1B = - 1
Putting above values in equation (1), we will get,
∴1(1−t)(2−t)=11−t−12−t\therefore \dfrac{1}{{(1 - t)(2 - t)}} = \dfrac{1}{{1 - t}} - \dfrac{1}{{2 - t}}
Ingratiating both sides, we will get,
∴∫1(1−t)(2−t)dt=∫dt1−t+∫dtt−2\therefore \int {\dfrac{1}{{(1 - t)(2 - t)}}dt = \int {\dfrac{{dt}}{{1 - t}} + \int {\dfrac{{dt}}{{t - 2}}} } }
∴∫dt(1−t)(2−t)=−In(1−t)+In(t−2)+c\therefore \int {\dfrac{{dt}}{{(1 - t)(2 - t)}} = - In(1 - t) + In(t - 2)} + c
Now, putting t=cos⁡xt = \cos x, we will get,
∴∫sin⁡x(1−cos⁡x)(2−cos⁡x)dx=In(cos⁡x−21−cos⁡x)+c\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.