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$\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{2}{3}} x}{\sin ^{\frac{2}{3}} x+\cos ^{\frac{2}{3}} x} d x=$
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Verified Answer
The correct answer is:
$\frac{\pi}{4}$
$\begin{aligned} I &=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{2}{3}} x}{\sin ^{\frac{2}{3}} x+\cos ^{\frac{2}{3}} x} d x ...(1)\\ &=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{2}{3}}\left(\frac{\pi}{2}-x\right)}{\sin ^{\frac{2}{3}}\left(\frac{\pi}{2}-x\right)+\cos ^{\frac{2}{3}}\left(\frac{\pi}{2}-x\right)} d x \\ \therefore I &=\int_{0}^{2} \frac{\cos ^{\frac{2}{3}} x}{\sin ^{\frac{2}{3}} x+\cos ^{\frac{2}{3}} x} d x ...(2) \end{aligned}$
Adding equation (1) \& (2) we get
$\begin{aligned}
2 I &=\int_{0}^{2} 1 d x \Rightarrow 2 I=[x]_{0}^{\frac{\pi}{2}} \\
I &=\frac{1}{2}\left(\frac{\pi}{2}-0\right)=\frac{\pi}{4}
\end{aligned}$
Adding equation (1) \& (2) we get
$\begin{aligned}
2 I &=\int_{0}^{2} 1 d x \Rightarrow 2 I=[x]_{0}^{\frac{\pi}{2}} \\
I &=\frac{1}{2}\left(\frac{\pi}{2}-0\right)=\frac{\pi}{4}
\end{aligned}$
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