<s>[INST] Solving the integral of $\cos^2x\sin^2 x$ Solving the integral of $\cos^2x\sin^2 x$:
My steps are: $(\cos x\sin x)^2=\left(\frac{\sin(2x)}{2}\right)^2$. Now we know that
$$\sin^2(\alpha)=\frac{1-\cos (2x)}{2}\iff\left(\frac{\sin(2x)}{2}\right)^2=\frac 14\sin^2(2x)=\frac 14\cdot \frac{1-\cos (4x)}{2}=\frac 18(1-\cos (4x))$$
Rewriting all the steps:
$$\int \sin ^2\left(x\right)\cos ^2\left(x\right)dx=\frac{1}{8}\left(x-\frac{1}{4}\sin \left(4x\right)\right)+k, \,\, k\in \Bbb R \tag 1$$
Is there another method to solve this integral (1)?
 [/INST] Integrate by parts
\begin{align}
\int \cos^2x\sin^2 x \ dx =& -\frac1{16}\int\tan 2x\ d(\cos^2 2x)= -\frac1{16}\tan2x\cos^22x+\frac18 x+C
\end{align}
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