Q114 / 160 · TS EAMCET 2013 · Mathematics
If
$$
\int \frac{x-\sin x}{1+\cos x} d x=x \tan \left(\frac{x}{2}\right)+p \log \left|\sec \left(\frac{x}{2}\right)\right|+C,
$$
then $p$ is equal to
- Correct answerA$-4$
- B$4$
- C$2$
- D$-2$
Solution
Let $\begin{aligned} I & =\int \frac{x-\sin x}{1+\cos x} d x \\ & =\int \frac{x}{1+\cos x} d x-\int \frac{\sin x}{1+\cos x} d x \\ & =\frac{1}{2} \int \frac{x}{\cos ^2\left(\frac{x}{2}\right)} d x-\int \frac{2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}{2 \cos ^2\left(\frac{x}{2}\right)} d x \\ & =\frac{1}{2} \int x \sec ^2\left(\frac{x}{2}\right) d x-\int \tan \left(\frac{x}{2}\right) d x \\ & =\frac{1}{2}\left\{x \cdot 2 \tan \left(\frac{x}{2}\right)-\int 2 \tan \left(\frac{x}{2}\right) d x\right\} \\ & =x \tan \frac{x}{2}-\int \tan \frac{x}{2} d x-\int \tan \left(\frac{x}{2}\right)+C \\ & =x \tan \frac{x}{2}-4 \log \left|\sec \frac{x}{2}\right|+C\end{aligned}$
but given,
$$
\int \frac{x-\sin x}{1+\cos x} d x=x \tan \frac{x}{2}+p \log \left|\sec \frac{x}{2}\right|+C
$$
On comparing, we get
$$
p=-4
$$
Hence, the correct answer is option: 'A'.