int_{0}^{frac{pi}{2}} frac{sqrt[3]{sec x}}{sq | vedclass

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Question

(B) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{(\sec x)^{1/3}}{(\sec x)^{1/3} + (\operatorname{cosec} x)^{1/3}} dx \quad ...(1)$ Using the property $\int

Vedclass · Accepted Answer

$\frac{\pi}{4}$

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(B) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{(\sec x)^{1/3}}{(\sec x)^{1/3} + (\operatorname{cosec} x)^{1/3}} dx \quad ...(1)$ Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we have: $I = \int_{0}^{\frac{\pi}{2}} \frac{(\sec(\frac{\pi}{2}-x))^{1/3}}{(\sec(\frac{\pi}{2}-x))^{1/3} + (\operatorname{cosec}(\frac{\pi}{2}-x))^{1/3}} dx$ Since $\sec(\frac{\pi}{2}-x) = \operatorname{cosec} x$ and $\operatorname{cosec}(\frac{\pi}{2}-x) = \sec x$,we get: $I = \int_{0}^{\frac{\pi}{2}} \frac{(\operatorname{cosec} x)^{1/3}}{(\operatorname{cosec} x)^{1/3} + (\sec x)^{1/3}} dx \quad ...(2)$ Adding equation $(1)$ and $(2)$: $2I = \int_{0}^{\frac{\pi}{2}} \frac{(\sec x)^{1/3} + (\operatorname{cosec} x)^{1/3}}{(\sec x)^{1/3} + (\operatorname{cosec} x)^{1/3}} dx$ $2I = \int_{0}^{\frac{\pi}{2}} 1 dx = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2}$ Therefore,$I = \frac{\pi}{4}$.

$\int_{0}^{\frac{\pi}{2}} \frac{\sqrt[3]{\sec x}}{\sqrt[3]{\sec x}+\sqrt[3]{\operatorname{cosec} x}} d x=$

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