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$\int_{0}^{\frac{\pi}{2}} \frac{\sqrt[7]{\sin x}}{\sqrt[7]{\sin x}+\sqrt[7]{\cos x}} d x=$
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Verified Answer
The correct answer is:
$\frac{\pi}{4}$
(D)
$I=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt[7]{\sin x}}{\sqrt[7]{\sin x}+\sqrt[7]{\cos x}}$...(1)
$=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)+\sqrt{\cos \left(\frac{\pi-x}{2}\right)}} \mathrm{dx}}$
$I=\int_{0}^{r / 2} \frac{\sqrt[7]{\cos x}}{\sqrt[7]{\cos x}+\sqrt[7]{\sin x}} d x$...(2)
Equation $(1)-(2)$ gives
$\begin{aligned}
2 I &=\int_{0}^{\frac{\pi}{2}} 1 d x=[x]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2} \\
\therefore I &=\frac{\pi}{4}
\end{aligned}$
$I=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt[7]{\sin x}}{\sqrt[7]{\sin x}+\sqrt[7]{\cos x}}$...(1)
$=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)+\sqrt{\cos \left(\frac{\pi-x}{2}\right)}} \mathrm{dx}}$
$I=\int_{0}^{r / 2} \frac{\sqrt[7]{\cos x}}{\sqrt[7]{\cos x}+\sqrt[7]{\sin x}} d x$...(2)
Equation $(1)-(2)$ gives
$\begin{aligned}
2 I &=\int_{0}^{\frac{\pi}{2}} 1 d x=[x]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2} \\
\therefore I &=\frac{\pi}{4}
\end{aligned}$
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