Question

Evaluate the value of $\int {\dfrac{{x - \sin x}}{{1 - \cos x}}dx} $ .

Answer 1

Hint: Here, we are asked to find the value of $\int {\dfrac{{x - \sin x}}{{1 - \cos x}}dx} $ .
Let $I = \int {\dfrac{{x - \sin x}}{{1 - \cos x}}dx} $ and put $\cos x = 1 - 2{\sin ^2}\dfrac{x}{2}$ and $\sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$ .
Then, solve the integral by spitting it in two parts ${I_1}$ and ${I_2}$ .
Thus, find the required answer of given integral.

Complete step-by-step answer:
Here, we are asked to find the value of $\int {\dfrac{{x - \sin x}}{{1 - \cos x}}dx} $ .
Let $I = \int {\dfrac{{x - \sin x}}{{1 - \cos x}}dx} $ .
Now, $\cos x = 1 - 2{\sin ^2}\dfrac{x}{2}$
 \[
  \therefore I = \int {\dfrac{{x - \sin x}}{{1 - \left( {1 - 2{{\sin }^2}\dfrac{x}{2}} \right)}}dx} \\
  \therefore I = \int {\dfrac{{x - \sin x}}{{1 - 1 + 2{{\sin }^2}\dfrac{x}{2}}}dx} \\
  \therefore I = \int {\dfrac{{x - \sin x}}{{2{{\sin }^2}\dfrac{x}{2}}}dx} \\
  \therefore I = \dfrac{1}{2}\int {\dfrac{x}{{{{\sin }^2}\dfrac{x}{2}}}dx} - \dfrac{1}{2}\int {\dfrac{{\sin x}}{{{{\sin }^2}\dfrac{x}{2}}}dx} \\
 \]
Also, $\sin x$ can be written as $2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$
 $
  \therefore I = \dfrac{1}{2}\int {x{{\operatorname{cosec} }^2}\dfrac{x}{2}dx} - \dfrac{1}{2}\int {\dfrac{{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}{{{{\sin }^2}\dfrac{x}{2}}}dx} \\
  \therefore I = \dfrac{1}{2}\int {x{{\operatorname{cosec} }^2}\dfrac{x}{2}dx} - \dfrac{2}{2}\int {\dfrac{{\cos \dfrac{x}{2}}}{{\sin \dfrac{x}{2}}}dx} \\
  \therefore I = \dfrac{1}{2}\int {x{{\operatorname{cosec} }^2}\dfrac{x}{2}dx} - \int {\cot \dfrac{x}{2}dx} \\
 $
Let, ${I_1} = \int {x{{\operatorname{cosec} }^2}\dfrac{x}{2}dx} $ and ${I_2} = \int {\cot \dfrac{x}{2}dx} $
First, we will solve ${I_1}$ .
 $\therefore {I_1} = \int {x{{\operatorname{cosec} }^2}\dfrac{x}{2}dx} $
 $
  \therefore {I_1} = x\int {{{\operatorname{cosec} }^2}\dfrac{x}{2}dx} - \int {\left( {\dfrac{{dx}}{{dx}} \cdot \int {{{\operatorname{cosec} }^2}\dfrac{x}{2}dx} } \right)dx} \\
  \therefore {I_1} = x\left( { - 2\cot \dfrac{x}{2}} \right) - \int { - 2\cot \dfrac{x}{2}dx} \\
  \therefore {I_1} = - 2x\cot \dfrac{x}{2} + 2\int {\cot \dfrac{x}{2}dx} \\
  \therefore {I_1} = - 2x\cot \dfrac{x}{2} + 2\ln \left| {\sin \dfrac{x}{2} \times 2} \right| \\
 $
Then, we will solve ${I_2}$ .
 $
  \therefore {I_2} = \int {\cot \dfrac{x}{2}dx} \\
  \therefore {I_2} = \ln \left| {\sin \dfrac{x}{2} \times 2} \right| \\
 $
Thus, $I = \dfrac{1}{2}{I_1} - {I_2}$
 $
  \therefore I = \dfrac{1}{2}\left[ { - 2x\cot \dfrac{x}{2} + 2\ln \left| {2\sin \dfrac{x}{2}} \right|} \right] - \ln \left| {2\sin \dfrac{x}{2}} \right| + C \\
  \therefore I = - x\cot \dfrac{x}{2} + \ln \left| {2\sin \dfrac{x}{2}} \right| - \ln \left| {2\sin \dfrac{x}{2}} \right| + C \\
  \therefore I = - x\cot \dfrac{x}{2} + C \\
 $
Thus, $\int {\dfrac{{x - \sin x}}{{1 - \cos x}}dx} = - x\cot \dfrac{x}{2} + C$.

Note: In these types of questions, we may not see the answer until the end as the terms cancel out in the end, Hence we should be patient enough to solve these types of questions. Also while doing the substitutions, we must be very careful on the terms dx and dt, as there are high chances that we forgot to change from dx to dt.