Evaluate the definite integral int_{0}^{frac{ | vedclass

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(B) Let $I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x$. Divide the numerator and denominator by $\cos^{4} x$: $I = \

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$\frac{\pi}{8}$

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(B) Let $I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x$. Divide the numerator and denominator by $\cos^{4} x$: $I = \int_{0}^{\frac{\pi}{4}} \frac{	an x \sec^{2} x}{1+	an^{4} x} d x$. Let $	an^{2} x = t$. Then $2 	an x \sec^{2} x d x = d t$,which implies $	an x \sec^{2} x d x = \frac{1}{2} d t$. When $x = 0$,$t = 	an^{2} 0 = 0$. When $x = \frac{\pi}{4}$,$t = 	an^{2} \frac{\pi}{4} = 1$. Substituting these into the integral: $I = \frac{1}{2} \int_{0}^{1} \frac{d t}{1+t^{2}}$. Using the standard integral $\int \frac{1}{1+t^{2}} d t = 	an^{-1} t$: $I = \frac{1}{2} [	an^{-1} t]_{0}^{1} = \frac{1}{2} (	an^{-1} 1 - 	an^{-1} 0) = \frac{1}{2} (\frac{\pi}{4} - 0) = \frac{\pi}{8}$.

Evaluate the definite integral $\int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x$.

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