Question

${\int }_0^{\frac{\pi }{2}}\frac{cosx}{1+cosx+sinx}dx=$

• A. $\frac{\pi }{4}+\frac{1}{2}log2$
• B. $\frac{\pi }{4}+log2$
• C. $\frac{\pi }{4}-\frac{1}{2}log2$
• D. $\frac{\pi }{4}-log2$

Answer 1

The correct option is C $\frac{\pi }{4}-\frac{1}{2}log2$

${\int }_0^{\frac{\pi }{2}}\frac{cosx}{1+cosx+sinx}dx$
$={\int }_0^{\frac{\pi }{2}}\frac{co{s}^2\left(\frac{x}{2}\right)-si{n}^2\left(\frac{x}{2}\right)}{2co{s}^2\left(\frac{x}{2}\right)+2sin\left(\frac{x}{2}\right)cos\left(\frac{x}{2}\right)}dx$
$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{1-ta{n}^2\left(\frac{x}{2}\right)}{1+tan\left(\frac{x}{2}\right)}dx=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}[1-tan\left(\frac{x}{2}\right)]dx$
$\frac{\pi }{4}+log\frac{1}{\sqrt{2}}=\frac{\pi }{4}-\frac{1}{2}log2$