int_0^{frac{pi}{4}} frac{cos^2 x sin^2 x}{(co | vedclass

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

Vedclass · Accepted Answer

$\frac{1}{6}$

Vedclass · Answer

(C) Let $I = \int_0^{\frac{\pi}{4}} \frac{\cos^2 x \sin^2 x}{(\cos^3 x + \sin^3 x)^2} \, dx$. Divide the numerator and denominator by $\cos^6 x$: $I = \int_0^{\frac{\pi}{4}} \frac{	an^2 x \sec^2 x}{(1 + 	an^3 x)^2} \, dx$. Let $u = 	an^3 x$,then $du = 3 	an^2 x \sec^2 x \, dx$,which implies $	an^2 x \sec^2 x \, dx = \frac{du}{3}$. When $x = 0$,$u = 0$. When $x = \frac{\pi}{4}$,$u = 1$. Substituting these into the integral: $I = \int_0^1 \frac{1}{3(1 + u)^2} \, du = \frac{1}{3} \left[ -\frac{1}{1 + u} ight]_0^1$. $I = \frac{1}{3} \left( -\frac{1}{2} - (-1) ight) = \frac{1}{3} \left( \frac{1}{2} ight) = \frac{1}{6}$.

$\int_0^{\frac{\pi}{4}} \frac{\cos^2 x \sin^2 x}{(\cos^3 x + \sin^3 x)^2} \, dx =$

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