int_{0}^{frac{pi}{2}} frac{sqrt[7]{sin x}}{sq | vedclass

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(C) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt[7]{\sin x}}{\sqrt[7]{\sin x}+\sqrt[7]{\cos x}} dx$ ... $(1)$Using the property $\int_{0}^{a} f(x) dx

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$\frac{\pi}{4}$

Vedclass · Answer

(C) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt[7]{\sin x}}{\sqrt[7]{\sin x}+\sqrt[7]{\cos x}} dx$ ... $(1)$Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get:$I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt[7]{\sin(\frac{\pi}{2}-x)}}{\sqrt[7]{\sin(\frac{\pi}{2}-x)}+\sqrt[7]{\cos(\frac{\pi}{2}-x)}} dx$$I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt[7]{\cos x}}{\sqrt[7]{\cos x}+\sqrt[7]{\sin x}} dx$ ... $(2)$Adding equations $(1)$ and $(2)$:$2I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt[7]{\sin x} + \sqrt[7]{\cos x}}{\sqrt[7]{\sin x} + \sqrt[7]{\cos x}} dx$$2I = \int_{0}^{\frac{\pi}{2}} 1 dx$$2I = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2}$$I = \frac{\pi}{4}$

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

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