Question

Solve $I$ = $\int \frac{x+\sin x}{1+\cos x}dx$ then,

• A. $I$ = $x\tan \frac{x}{2}+c$
• B. $I$ = $x{\sec }^2\frac{x}{2}+c$
• C. $I$ = $\log \cos \frac{x}{2}$
• D. None of these

Answer 1

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

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

$I=\int \frac{x+\sin x}{2{\cos }^2\frac{x}{2}}dx=\frac{1}{2}\int x{\sec }^2\frac{x}{2}dx+\int \frac{2\sin \frac{x}{2}\cos \frac{x}{2}}{2{\cos }^2\frac{x}{2}dx}$

$I=\frac{1}{2}[x\int {\sec }^2\frac{x}{2}-\int \{\frac{d}{dx}x\int {\sec }^2\frac{x}{2}dx\}]+\int \tan \frac{x}{2}dx$

$I=\frac{1}{2}[x.2.\tan \frac{x}{2}-\int 2.\tan \frac{x}{2}dx]+2ln\left|\sec \frac{x}{2}\right|+c$

• $I=\frac{1}{2}[2x\tan \frac{x}{2}-4ln\left|\sec \frac{x}{2}\right|]+2ln\left|\sec \frac{x}{2}\right|+c$
  

$I=x\tan \frac{x}{2}-2ln|\sec \frac{x}{2}|+2ln|\sec \frac{x}{2}|+c$

$I=x\tan \frac{x}{2}+c$