Evaluate int_{0}^{frac{pi}{2}} frac{sin ^{4} | vedclass

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(B) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x$ ..... $(1)$Using the property $\int_{0}^{a} f(x) dx = \int_{0}^

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$\frac{\pi}{4}$

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

Evaluate $\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x$

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