Question

${\int }_0^{\frac{\pi }{2}}\frac{\cos xdx}{\cos x+\sin x}$

Answer 1

We have,

$I={\int }_0^{\frac{\pi }{2}}\frac{\cos xdx}{\cos x+\sin x}$ $......\left(1\right)$

On applying property,

$I={\int }_0^{\frac{\pi }{2}}\frac{\cos (\frac{\pi }{2}-x)dx}{\cos (\frac{\pi }{2}-x)+\sin (\frac{\pi }{2}-x)}$

$I={\int }_0^{\frac{\pi }{2}}\frac{\sin xdx}{\sin x+\cos x}$ $......\left(2\right)$

On adding equations $\left(1\right)$ and $\left(2\right)$, we get

$2I={\int }_0^{\frac{\pi }{2}}\frac{\cos x+\sin x}{\cos x+\sin x}dx$

$2I={\int }_0^{\frac{\pi }{2}}1dx$

$2I=[x{]}_0^{\frac{\pi }{2}}$

$2I=\frac{\pi }{2}-0$

$2I=\frac{\pi }{2}$

$I=\frac{\pi }{4}$

Hence, this is the answer.