Correct Answer - Option 1 :
\(\rm \;\log \left| {\dfrac{{1 - \cos x}}{{2 - \cos x}}} \right| + C\)
Concept:
Partial Fraction:
|
Factors in the denominator
|
Corresponding Partial Fraction
|
|
(x - a)
|
\(\dfrac{A}{{x - a}}\)
|
|
(x – b)2
|
\(\dfrac{A}{{x - b}} + \dfrac{B}{{{{\left( {x - b} \right)}^2}}}\)
|
|
(x - a) (x – b)
|
\(\dfrac{A}{{\left( {x - a} \right)}} + \dfrac{B}{{\left( {x - b} \right)}}\)
|
|
(x – c)3
|
\(\dfrac{A}{{x - c}} + \dfrac{B}{{{{\left( {x - c} \right)}^2}}} + \dfrac{C}{{{{\left( {x - c} \right)}^3}}}\)
|
|
(x – a) (x2 – a)
|
\(\dfrac{A}{{\left( {x - a} \right)}} + \dfrac{{Bx + C}}{{\left( {{x^2} - a} \right)}}\)
|
|
(ax2 + bx + c)
|
\(\dfrac{{Ax + B}}{{\left( {a{x^2} + bx + c} \right)}}\)
|
Calculation:
Given: \(\smallint \dfrac{{\sin x}}{{\left( {1 -\ \cos x} \right)\;\left( {2 -\ \cos x} \right)}}dx\)
Let cos x = t and - sin dx = dt
\(\smallint \dfrac{{\sin x}}{{\left( {1 - \cos x} \right)\;\left( {2 - \cos x} \right)}}dx = -\;\smallint \dfrac{{dt}}{{\left( {1 - t} \right)\;\left( {2 - t} \right)}}\)
Let \(\dfrac{1}{{\left( {1 - t} \right)\;\left( {2 - t} \right)}} = \dfrac{A}{{1 - t}} + \dfrac{B}{{2 - t}}\)
⇒ 1 = A (2 - t) + B (1 - t) ---(1)
By putting t = 1 on both the sides of (1) we get A = 1
By putting t = 2 on both the sides of (1) we get B = - 1
\(\Rightarrow \dfrac{1}{{\left( {1 - t} \right)\;\left( {2 - t} \right)}} = \dfrac{1}{{1 - t}} - \dfrac{1}{{2 - t}}\)
\(\Rightarrow -\smallint \dfrac{{dt}}{{\left( {1 - t} \right)\;\left( {2 - t} \right)}} = -\left[\;\smallint \dfrac{{dt}}{{1 - t}} - \;\smallint \dfrac{{dt}}{{2 - t}}\right]\)
\(\rm \Rightarrow \;\smallint \dfrac{{dt}}{{2 - t}} -\;\smallint \dfrac{{dt}}{{1 - t}} \)
As we know that \(\smallint \dfrac{{dx}}{x} = \log \left| x \right|\; + C\) where C is a constant
\(\Rightarrow \smallint \dfrac{{dt}}{{\left( {1 - t} \right)\;\left( {2 - t} \right)}} = \; - \log \left| {2 - t} \right| + \log \left| {1 - t} \right| + C\)
\(= \log \left| {\dfrac{{1 - t}}{{2 - t}}} \right| + C\)
By substituting cos x = t in the above equation we get,
\(\Rightarrow \smallint \dfrac{{\cos x}}{{\left( {1 - \sin x} \right)\;\left( {2 - \sin x} \right)}}dx = \;\log \left| {\dfrac{{1 - \cos x}}{{2 - \cos x}}} \right| + C\)
