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

Evaluate the value of ∫x−sin⁡x1−cos⁡xdx\int {\dfrac{{x - \sin x}}{{1 - \cos x}}dx} .

Explanation

Solution

Here, we are asked to find the value of ∫x−sin⁡x1−cos⁡xdx\int {\dfrac{{x - \sin x}}{{1 - \cos x}}dx} .
Let I=∫x−sin⁡x1−cos⁡xdxI = \int {\dfrac{{x - \sin x}}{{1 - \cos x}}dx} and put cos⁡x=1−2sin⁡2x2\cos x = 1 - 2{\sin ^2}\dfrac{x}{2} and sin⁡x=2sin⁡x2cos⁡x2\sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} .
Then, solve the integral by spitting it in two parts I1{I_1} and I2{I_2} .
Thus, find the required answer of given integral.

Complete step-by-step answer:
Here, we are asked to find the value of ∫x−sin⁡x1−cos⁡xdx\int {\dfrac{{x - \sin x}}{{1 - \cos x}}dx} .
Let I=∫x−sin⁡x1−cos⁡xdxI = \int {\dfrac{{x - \sin x}}{{1 - \cos x}}dx} .
Now, cos⁡x=1−2sin⁡2x2\cos x = 1 - 2{\sin ^2}\dfrac{x}{2}

∴I=∫x−sin⁡x1−(1−2sin⁡2x2)dx ∴I=∫x−sin⁡x1−1+2sin⁡2x2dx ∴I=∫x−sin⁡x2sin⁡2x2dx ∴I=12∫xsin⁡2x2dx−12∫sin⁡xsin⁡2x2dx  \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\sin x can be written as 2sin⁡x2cos⁡x22\sin \dfrac{x}{2}\cos \dfrac{x}{2}
∴I=12∫xcosec⁡2x2dx−12∫2sin⁡x2cos⁡x2sin⁡2x2dx ∴I=12∫xcosec⁡2x2dx−22∫cos⁡x2sin⁡x2dx ∴I=12∫xcosec⁡2x2dx−∫cot⁡x2dx  \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, I1=∫xcosec⁡2x2dx{I_1} = \int {x{{\operatorname{cosec} }^2}\dfrac{x}{2}dx} and I2=∫cot⁡x2dx{I_2} = \int {\cot \dfrac{x}{2}dx}
First, we will solve I1{I_1} .
∴I1=∫xcosec⁡2x2dx\therefore {I_1} = \int {x{{\operatorname{cosec} }^2}\dfrac{x}{2}dx}
∴I1=x∫cosec⁡2x2dx−∫(dxdx⋅∫cosec⁡2x2dx)dx ∴I1=x(−2cot⁡x2)−∫−2cot⁡x2dx ∴I1=−2xcot⁡x2+2∫cot⁡x2dx ∴I1=−2xcot⁡x2+2ln⁡∣sin⁡x2×2∣  \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 I2{I_2} .
∴I2=∫cot⁡x2dx ∴I2=ln⁡∣sin⁡x2×2∣  \therefore {I_2} = \int {\cot \dfrac{x}{2}dx} \\\ \therefore {I_2} = \ln \left| {\sin \dfrac{x}{2} \times 2} \right| \\\
Thus, I=12I1−I2I = \dfrac{1}{2}{I_1} - {I_2}
∴I=12[−2xcot⁡x2+2ln⁡∣2sin⁡x2∣]−ln⁡∣2sin⁡x2∣+C ∴I=−xcot⁡x2+ln⁡∣2sin⁡x2∣−ln⁡∣2sin⁡x2∣+C ∴I=−xcot⁡x2+C  \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, ∫x−sin⁡x1−cos⁡xdx=−xcot⁡x2+C\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.