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

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(D) Let $I = \int_{-\pi}^{\pi} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx$. Since the integrand $f(x) = \frac{\sin^4 x}{\sin^4 x + \cos^4 x}$ is an ev

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$\pi$

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(D) Let $I = \int_{-\pi}^{\pi} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx$. Since the integrand $f(x) = \frac{\sin^4 x}{\sin^4 x + \cos^4 x}$ is an even function,we have $I = 2 \int_{0}^{\pi} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx$. Using the property $\int_{0}^{2a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$ if $f(2a-x) = f(x)$,we note that $f(\pi - x) = \frac{\sin^4(\pi - x)}{\sin^4(\pi - x) + \cos^4(\pi - x)} = \frac{\sin^4 x}{\sin^4 x + \cos^4 x} = f(x)$. Thus,$I = 2 	imes 2 \int_{0}^{\pi/2} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx = 4 \int_{0}^{\pi/2} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx$ ..... $(i)$. Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we get $I = 4 \int_{0}^{\pi/2} \frac{\cos^4 x}{\cos^4 x + \sin^4 x} \, dx$ ..... $(ii)$. Adding $(i)$ and $(ii)$,we get $2I = 4 \int_{0}^{\pi/2} \frac{\sin^4 x + \cos^4 x}{\sin^4 x + \cos^4 x} \, dx = 4 \int_{0}^{\pi/2} 1 \, dx = 4 	imes \frac{\pi}{2} = 2\pi$. Therefore,$I = \pi$.

$\int_{-\pi}^{\pi} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx = $

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