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

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Question

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

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

Vedclass · Answer

(B) Let $I = \int \frac{\sin ^2 x \cos ^2 x \,dx}{(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x)^2}$.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,$I = \int \frac{\sin ^2 x \cos ^2 x \,dx}{(\sin ^3 x+\cos ^3 x)^2}$.Divide the numerator and denominator by $\cos ^6 x$: $I = \int \frac{	an ^2 x \sec ^2 x \,dx}{(	an ^3 x+1)^2}$.Let $u = 	an ^3 x+1$,then $du = 3 	an ^2 x \sec ^2 x \,dx$,so $	an ^2 x \sec ^2 x \,dx = \frac{du}{3}$.Substituting these into the integral: $I = \int \frac{1}{u^2} \cdot \frac{du}{3} = \frac{1}{3} \int u^{-2} \,du = \frac{1}{3} \left(\frac{u^{-1}}{-1}ight) + C = -\frac{1}{3u} + C$.Substituting back $u = 	an ^3 x+1$: $I = \frac{-1}{3(1+	an ^3 x)}+C$.

The integral $\int \frac{\sin ^2 x \cos ^2 x \,dx}{(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x)^2}$ is equal to (where $C$ is a constant of integration).

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