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)?

You can excpress $\cos^2 x$ as $1 - \sin^2 x$ and hence your primitive becomes:
$$\int \sin^2 x d x - \int \sin^4 x dx$$
which comes to the same answer in the end.
Full working:
$$\int \sin^2 a x \cos^2 a x \, \mathrm d x$$
$$=\int \sin^2 a x (1 - \sin^2 a x) \, \mathrm d x$$
$$=\int \sin^2 a x \, \mathrm d x - \int \sin^4 a x \, \mathrm d x$$
$$=\frac x 2 - \frac {\sin 2 a x} {4 a} - \int \sin^4 a x \, \mathrm d x + C$$
$$= \frac x 2 - \frac {\sin 2 a x} {4 a} - \left ({\frac {3 x} 8 - \frac {\sin 2 a x} {4 a} + \frac {\sin 4 a x} {32 a} }\right) + C$$
$$= \frac x 8 - \frac {\sin 4 a x} {32 a} + C$$
In the above:
$$\int \sin^4 a x \, \mathrm d x = \int \left( {\frac {3 - 4 \cos 2 a x + \cos 4 a x} 8}\right) \, \mathrm d x$$
using the usual power reduction formula.
