Q81 / 150 · MHT CET 2021 (20 Sep Shift 1) · Mathematics

$\int \frac{x+\sin x}{1+\cos x} d x=$

  • Correct answer
    A
    $x \tan \left(\frac{x}{2}\right)+c$
  • B
    $\log (x+\sin x)+c$
  • C
    $\cot \left(\frac{x}{2}\right)+c$
  • D
    $\log (1+\cos x)+c$

Solution

$\begin{aligned} & \text { Let } I=\int \frac{x+\sin x}{1+\cos x} d x=\int \frac{x+\sin x}{2 \cos ^2 \frac{x}{2}} \\ & =\int \frac{x}{2 \cos ^2 \frac{x}{2}} d x+\int \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{\cos ^2 \frac{x}{2}} d x=\frac{1}{2} \int x \sec ^2 \frac{x}{2} d x+\int \tan \frac{x}{2} d x \\ & =\frac{1}{2}\left[x \tan \frac{x}{2}(2)-\int 2 \tan \frac{x}{2} d x\right]-2 \log \left|\cos \frac{x}{2}\right|+c \\ & =x \tan \frac{x}{2}+2 \log \left|\cos \frac{x}{2}\right|-2 \log \left|\cos \frac{x}{2}\right|+c=\tan \frac{x}{2}+c\end{aligned}$

Hence, the correct answer is option: 'A'.

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