The integral int frac{sin ^2 x cos ^2 x}{left | vedclass

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Question

(B) Let $I = \int \frac{\sin ^2 x \cos ^2 x}{(\sin ^5 x + \cos ^3 x \sin ^2 x + \sin ^3 x \cos ^2 x + \cos ^5 x)^2} \,dx$. Factor the denominator: $\s

Vedclass · Accepted Answer

$\frac{-1}{3\left(1+	an ^3 xight)}+c$, where $c$ is a constant of integration.

Vedclass · Answer

(B) Let $I = \int \frac{\sin ^2 x \cos ^2 x}{(\sin ^5 x + \cos ^3 x \sin ^2 x + \sin ^3 x \cos ^2 x + \cos ^5 x)^2} \,dx$. Factor the denominator: $\sin ^5 x + \sin ^3 x \cos ^2 x + \cos ^3 x \sin ^2 x + \cos ^5 x = \sin ^3 x(\sin ^2 x + \cos ^2 x) + \cos ^3 x(\sin ^2 x + \cos ^2 x) = \sin ^3 x + \cos ^3 x$. Thus, $I = \int \frac{\sin ^2 x \cos ^2 x}{(\sin ^3 x + \cos ^3 x)^2} \,dx$. Divide the numerator and denominator by $\cos ^6 x$: $I = \int \frac{	an ^2 x \sec ^2 x}{(1 + 	an ^3 x)^2} \,dx$. Let $t = 1 + 	an ^3 x$. Then $dt = 3 	an ^2 x \sec ^2 x \,dx$, which implies $	an ^2 x \sec ^2 x \,dx = \frac{1}{3} dt$. Substituting these into the integral: $I = \frac{1}{3} \int \frac{1}{t^2} \,dt = \frac{1}{3} (-t^{-1}) + c = -\frac{1}{3t} + c$. Substituting $t$ back: $I = \frac{-1}{3(1 + 	an ^3 x)} + c$.

The integral $\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x\right)^2} \,d x$ is equal to

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