$\int_{0}^{\frac{\pi}{2}} \frac{\sqrt[7]{\sin x}}{\sqrt[7]{\sin x}+\sqrt[7]{\cos x}} d x=$

Question

Aicharya · Accepted Answer

$I=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt[7]{\sin x}}{\sqrt[7]{\sin x}+\sqrt[7]{\cos x}}$...(1) $=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin \left(\frac{\pi}{2}-xight)}}{\sqrt{\sin \left(\frac{\pi}{2}-xight)+\sqrt{\cos \left(\frac{\pi-x}{2}ight)}} \mathrm{dx}}$ $I=\int_{0}^{r / 2} \frac{\sqrt[7]{\cos x}}{\sqrt[7]{\cos x}+\sqrt[7]{\sin x}} d x$...(2) Equation $(1)-(2)$ gives $\begin{aligned} 2 I &=\int_{0}^{\frac{\pi}{2}} 1 d x=[x]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2} \ 	herefore I &=\frac{\pi}{4} \end{aligned}$

$\int_{0}^{\frac{\pi}{2}} \frac{\sqrt[7]{\sin x}}{\sqrt[7]{\sin x}+\sqrt[7]{\cos x}} d x=$

Solution