int_0^{pi / 4} frac{cos ^2 x sin ^2 x}{left(c | vedclass

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Question

(C) Let $I = \int_0^{\pi / 4} \frac{\cos ^2 x \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} d x$.Divide the numerator and denominator by $\cos^6 x$:$

Vedclass · Accepted Answer

$1 / 6$

Vedclass · Answer

(C) Let $I = \int_0^{\pi / 4} \frac{\cos ^2 x \sin ^2 x}{\left(\cos ^3 x+\sin ^3 xight)^2} d x$.Divide the numerator and denominator by $\cos^6 x$:$I = \int_0^{\pi / 4} \frac{\frac{\cos ^2 x \sin ^2 x}{\cos^6 x}}{\left(\frac{\cos ^3 x+\sin ^3 x}{\cos^3 x}ight)^2} d x = \int_0^{\pi / 4} \frac{	an^2 x \sec^2 x}{(1+	an^3 x)^2} d x$.Let $t = 1 + 	an^3 x$. Then $dt = 3 	an^2 x \sec^2 x d x$,which implies $	an^2 x \sec^2 x d x = \frac{dt}{3}$.When $x = 0$,$t = 1 + 0 = 1$. When $x = \pi / 4$,$t = 1 + (1)^3 = 2$.Substituting these into the integral:$I = \frac{1}{3} \int_1^2 \frac{dt}{t^2} = \frac{1}{3} \left[ -\frac{1}{t} ight]_1^2 = \frac{1}{3} \left( -\frac{1}{2} - (-1) ight) = \frac{1}{3} \left( \frac{1}{2} ight) = \frac{1}{6}$.

$\int_0^{\pi / 4} \frac{\cos ^2 x \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} d x$ is equal to

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