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The integral $\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x\right)^2} \mathrm{~d} x$ is equal to
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Verified Answer
The correct answer is:
$\frac{-1}{3\left(1+\tan ^3 x\right)}+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
Let
$$
\begin{aligned}
\mathrm{I} & =\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x\right)^2} \mathrm{~d} x \\
& =\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^5 x+\sin ^3 x \cos ^2 x+\cos ^5 x \sin ^2 x+\cos ^5 x\right)^2} \mathrm{~d} x \\
& =\int \frac{\sin ^2 x \cos ^2 x}{\left[\sin ^3 x\left(\sin ^2 x+\cos ^2 x\right)+\cos ^3 x\left(\sin ^2 x+\cos ^2 x\right)\right]^2} \mathrm{~d} x \\
& =\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^3 x+\cos ^3 x\right)^2} \mathrm{~d} x \\
& =\int \frac{\sec ^2 x \tan ^2 x}{\left(1+\tan ^3 x\right)^2} \mathrm{~d} x
\end{aligned}
$$
[Dividing numerator and denominator by $\cos ^6 x$ ]
Let $1+\tan ^3 x=t$
Differentiating w.r.t. $x$, we get
$$
\begin{aligned}
& 3 \tan ^2 x \sec ^2 x \mathrm{~d} x=\mathrm{dt} \\
& \tan ^2 x \sec ^2 x \mathrm{~d} x=\frac{1}{3} \mathrm{dt}
\end{aligned}
$$
$$
\therefore \quad \tan ^2 x \sec ^2 x \mathrm{~d} x=\frac{1}{3} \mathrm{dt}
$$
$$
\begin{aligned}
\therefore \quad \mathrm{I} & =\frac{1}{3} \int \frac{1}{\mathrm{t}^2} \mathrm{dt} \\
& =\frac{-1}{3 \mathrm{t}}+\mathrm{c} \\
& =\frac{-1}{3\left(1+\tan ^3 x\right)}+\mathrm{c}
\end{aligned}
$$
$$
\begin{aligned}
\mathrm{I} & =\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x\right)^2} \mathrm{~d} x \\
& =\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^5 x+\sin ^3 x \cos ^2 x+\cos ^5 x \sin ^2 x+\cos ^5 x\right)^2} \mathrm{~d} x \\
& =\int \frac{\sin ^2 x \cos ^2 x}{\left[\sin ^3 x\left(\sin ^2 x+\cos ^2 x\right)+\cos ^3 x\left(\sin ^2 x+\cos ^2 x\right)\right]^2} \mathrm{~d} x \\
& =\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^3 x+\cos ^3 x\right)^2} \mathrm{~d} x \\
& =\int \frac{\sec ^2 x \tan ^2 x}{\left(1+\tan ^3 x\right)^2} \mathrm{~d} x
\end{aligned}
$$
[Dividing numerator and denominator by $\cos ^6 x$ ]
Let $1+\tan ^3 x=t$
Differentiating w.r.t. $x$, we get
$$
\begin{aligned}
& 3 \tan ^2 x \sec ^2 x \mathrm{~d} x=\mathrm{dt} \\
& \tan ^2 x \sec ^2 x \mathrm{~d} x=\frac{1}{3} \mathrm{dt}
\end{aligned}
$$
$$
\therefore \quad \tan ^2 x \sec ^2 x \mathrm{~d} x=\frac{1}{3} \mathrm{dt}
$$
$$
\begin{aligned}
\therefore \quad \mathrm{I} & =\frac{1}{3} \int \frac{1}{\mathrm{t}^2} \mathrm{dt} \\
& =\frac{-1}{3 \mathrm{t}}+\mathrm{c} \\
& =\frac{-1}{3\left(1+\tan ^3 x\right)}+\mathrm{c}
\end{aligned}
$$
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