The integral int frac{sin^2 x cos^2 x}{(sin^3 | vedclass

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(B) Let $I = \int \frac{\sin^2 x \cos^2 x}{(\sin^3 x + \cos^3 x)^2} dx$Divide numerator and denominator by $\cos^6 x$:$I = \int \frac{\frac{\sin^2 x \

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

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

The integral $\int \frac{\sin^2 x \cos^2 x}{(\sin^3 x + \cos^3 x)^2} dx$ is equal to

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