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