Question

The value of $\int \frac{x+\sin x}{1+\cos x}dx$ is
$($where $C$ is integration constant$)$

• A. $x\tan \frac{x}{2}+c$
• B. $\frac{x+\tan x}{2}+c$
• C. $x^2\tan \left(\frac{x}{2}\right)+c$
• D. $\frac{x}{2}\tan \left(\frac{x}{2}\right)+c$

Answer 1

The correct option is A $x\tan \frac{x}{2}+c$

$I=\int \frac{x+\sin x}{1+\cos x}dx=\int \left(\frac{x+\sin x}{2{\cos }^2\frac{x}{2}}\right)dx[\because 1+\cos 2x=2{\cos }^{2}x]=\int (x\cdot \frac{1}{2}{\sec }^2\frac{x}{2}+\tan \frac{x}{2})dx\underset{x{f}^{\prime }\left(x\right)}{\downarrow }\underset{f\left(x\right)}{\downarrow }[\because \sin x=2\sin \frac{x}{2}\cos \frac{x}{2}]=x\tan \frac{x}{2}+c[\because \int \left(f\right(x)+x{f}^{\prime }(x\left)\right)dx=x\cdot f\left(x\right)+c]$