int_{0}^{pi} frac{cos ^{4} x}{cos ^{4} x+sin | vedclass

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(B) Let $I = \int_{0}^{\pi} \frac{\cos ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x$. Since $f(\pi - x) = \frac{\cos^4(\pi - x)}{\sin^4(\pi - x) + \cos^4(\pi

Vedclass · Accepted Answer

$\frac{\pi}{2}$

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

$\int_{0}^{\pi} \frac{\cos ^{4} x}{\cos ^{4} x+\sin ^{4} x} d x$ is equal to

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