Question

$$\begin{gathered} \int \frac{x+\sin x}{1+coxx}dxisequalto \\ \left(\mathrm{a}\right)\log \left|1+\cos x\right|+C\left(\mathrm{b}\right)\log \left|x+\sin x\right|+C \\ \left(\mathrm{c}\right)x-\tan \frac{x}{2}+C\left(\mathrm{d}\right)x\tan \frac{x}{2}+C \end{gathered}$$

Answer 1

$\int \frac{x+\sin x}{1+\cos x}dx$

$=\int \frac{x+2\sin {\displaystyle \frac{x}{2}}\cos {\displaystyle \frac{x}{2}}}{2{\cos }^2{\displaystyle \frac{x}{2}}}dx$
$=\int \frac{x}{2{\cos }^2{\displaystyle \frac{x}{2}}}dx+\int \frac{2\sin {\displaystyle \frac{x}{2}}\cos {\displaystyle \frac{x}{2}}}{2{\cos }^2{\displaystyle \frac{x}{2}}}dx$

$=\frac{1}{2}\int x{\sec }^2\frac{x}{2}dx+\int \tan \frac{x}{2}dx$

$=\frac{1}{2}\left[x\int {\sec }^2\frac{x}{2}dx-\int \left(\frac{d}{dx}x\int {\sec }^2\frac{x}{2}dx\right)dx\right]+\int \tan \frac{x}{2}dx$

$=\frac{1}{2}\left[x\times \frac{\tan {\displaystyle \frac{x}{2}}}{{\displaystyle \frac{1}{2}}}-\int \left(\frac{\tan {\displaystyle \frac{x}{2}}}{{\displaystyle \frac{1}{2}}}\right)dx\right]+\int \tan \frac{x}{2}dx$

$=x\tan \frac{x}{2}-\int \tan \frac{x}{2}dx+\int \tan \frac{x}{2}dx+C$

$=x\tan \frac{x}{2}+C$

Hence, the correct answer is option (d).