Question

${\int }_0^{\frac{x}{2}}\frac{x+sinx}{1+cosx}dx=$ [Roorkee 1978]

• A. -log 2
• B. log 2
• C. $\frac{\pi }{2}$
• D. \N

Answer 1

The correct option is C $\frac{\pi }{2}$

${\int }_0^{\frac{x}{2}}\frac{x+sinx}{1+cosx}dx={\int }_0^{\frac{x}{2}}\frac{x+sinx}{2co{s}^2\frac{x}{2}}dx$
$=\frac{1}{2}{\int }_0^{\frac{x}{2}}xse{c}^2\frac{x}{2}dx+{\int }_0^{\frac{x}{2}}tan\frac{x}{2}dx$
$={\left|xtan\frac{x}{2}\right|}_0^{\frac{\pi }{2}}=\frac{\pi }{2}tan\frac{\pi }{4}=\frac{\pi }{2}$