Question

If ${\int }_0^{\frac{\pi }{2}}\frac{sinx}{1+sinx+cosx}dx=k$, then ${\int }_0^{\frac{\pi }{2}}\frac{dx}{1+sinx+cosx}$ is-

• A. $\frac{k}{2}$
• B. $\frac{\pi }{2}-k$
• C. $\frac{\pi }{2}-2k$
• D. $\frac{\pi }{2}+k$

Answer 1

The correct option is C

$\frac{\pi }{2}-2k$

${\int }_0^{\frac{\pi }{2}}\frac{sinx}{1+sinx+cosx}dx={\int }_0^{\frac{\pi }{2}}\frac{cosx}{1+sinx+cosx}dx=k\Rightarrow {\int }_0^{\frac{\pi }{2}}\left(\frac{1}{1+sinx+cosx}\right)dx=2k\Rightarrow {\int }_0^{\frac{\pi }{2}}\frac{dx}{1+sinx+cosx}=\frac{\pi }{2}-2k$