By using the properties of definite integrals | vedclass

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(B) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{\cos ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x$ ..... $(1)$Using the property $\int_{0}^{a} f(x) d x = \int_{0}

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$\frac{\pi}{4}$

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(B) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{\cos ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x$ ..... $(1)$Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,we get:$I = \int_{0}^{\frac{\pi}{2}} \frac{\cos ^{5}(\frac{\pi}{2}-x)}{\sin ^{5}(\frac{\pi}{2}-x)+\cos ^{5}(\frac{\pi}{2}-x)} d x$Since $\cos(\frac{\pi}{2}-x) = \sin x$ and $\sin(\frac{\pi}{2}-x) = \cos x$,we have:$I = \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{5} x}{\cos ^{5} x+\sin ^{5} x} d x$ ..... $(2)$Adding $(1)$ and $(2)$:$2I = \int_{0}^{\frac{\pi}{2}} \frac{\cos ^{5} x + \sin ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x$$2I = \int_{0}^{\frac{\pi}{2}} 1 d x$$2I = [x]_{0}^{\frac{\pi}{2}}$$2I = \frac{\pi}{2} - 0 = \frac{\pi}{2}$$I = \frac{\pi}{4}$

By using the properties of definite integrals,evaluate the integral $\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x$.

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