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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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The correct answer is:
$\frac{-1}{3\left(1+\tan ^3 x\right)}+C$
$\begin{aligned} & \int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x\right)^2} d x \\ & =\int \frac{\tan ^2 x \cdot \sec ^6 x d x}{\left(\tan ^5 x+\tan ^2 x+\tan ^3 x+1\right)^2} \text { [dividing } N^r \text { and } D^r \text { by } \cos ^{10} x \text { ] } \\ & =\int \frac{\tan ^2 x \cdot\left(\tan ^2 x+1\right)^2 \sec ^2 x d x}{\left(\tan ^2 x+1\right)^2\left(\tan ^3 x+1\right)^2} \\ & =\int \frac{\tan ^2 x \cdot \sec ^2 x d x}{\left(\tan ^3 x+1\right)^2}\end{aligned}$
$\begin{aligned} & =\frac{1}{3} \int \frac{\mathrm{d} t}{t^2} \\ & \left\{\text { Let } \tan ^3 x+1=t \text { i.e., } 3 \tan ^2 x \cdot \sec ^2 x \mathrm{~d} x=\mathrm{d} t\right\} \\ & =-\frac{1}{3 t}+c \\ & =\frac{-1}{3\left(\tan ^3 x+1\right)}+c\end{aligned}$
$\begin{aligned} & =\frac{1}{3} \int \frac{\mathrm{d} t}{t^2} \\ & \left\{\text { Let } \tan ^3 x+1=t \text { i.e., } 3 \tan ^2 x \cdot \sec ^2 x \mathrm{~d} x=\mathrm{d} t\right\} \\ & =-\frac{1}{3 t}+c \\ & =\frac{-1}{3\left(\tan ^3 x+1\right)}+c\end{aligned}$
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