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

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(D) Let $I = \int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x\right)^2} d x$.Factor the denominator:

Vedclass · Accepted Answer

$\frac{-1}{3\left(1+	an ^3 xight)}+C$

Vedclass · Answer

(D) Let $I = \int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 xight)^2} d x$.Factor the denominator: $\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x = \sin ^2 x(\sin ^3 x + \cos ^3 x) + \cos ^2 x(\sin ^3 x + \cos ^3 x) = (\sin ^2 x + \cos ^2 x)(\sin ^3 x + \cos ^3 x) = (\sin ^3 x + \cos ^3 x)$.Thus,$I = \int \frac{\sin ^2 x \cos ^2 x}{(\sin ^3 x + \cos ^3 x)^2} d x$.Divide the numerator and denominator by $\cos ^6 x$: $I = \int \frac{	an ^2 x \cdot \sec ^2 x}{(	an ^3 x + 1)^2} d x$.Let $t = 	an ^3 x + 1$,then $dt = 3 	an ^2 x \sec ^2 x d x$,so $	an ^2 x \sec ^2 x d x = \frac{dt}{3}$.Substituting these into the integral: $I = \int \frac{1}{t^2} \cdot \frac{dt}{3} = \frac{1}{3} \int t^{-2} dt = \frac{1}{3} \left( \frac{t^{-1}}{-1} ight) + C = -\frac{1}{3t} + C$.Substituting back $t = 	an ^3 x + 1$,we get $I = \frac{-1}{3(1 + 	an ^3 x)} + C$.

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 (Where $C$ is a constant of integration).

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