Q:

Trig integrals sin and cos

Need some help on sinxcosx integrals, having trouble with both are even powered
$\int {\sin(2x)}^4\cos(2x)^4dx$, not sure what to do, I realize I need half angle formulas but I'm stuck.

A:

Method $\#1:$
The idea is we need to utilize multiple/double angle formula for $$\displaystyle\int\sin^max\cos^nax dx$$ if $m,n$ both are even
Using $\sin2A=2\sin A\cos A,$
$\displaystyle\sin^42x\cdot\cos^42x=(\sin2x\cos2x)^4=\frac{\sin^44x}{2^4}$
Again using $\displaystyle\cos2B=1-2\sin^2B\iff2\sin^2B=1-\cos2B,$
$\displaystyle \sin^44x=\frac{(2\sin^24x)^2}{2^2}=\frac{(1-\cos8x)^2}4=\frac{1-2\cos8x+\cos^28x}4$
Now use $\displaystyle \cos2C=2\cos^2C-1\iff2\cos^2C=1+\cos2C$ for $\displaystyle\cos^28x$

If at least one of $m,n$ is odd say $m,$ we need to set $\cos ax=u$

Method $\#2:$
Using Euler Formula, $\displaystyle \cos y=\frac{e^{iy}+e^{-iy}}2,\sin y=\frac{e^{iy}-e^{-iy}}{2i}$
$$\sin^42x\cdot\cos^42x=\left(\frac{e^{2ix}-e^{-2ix}}{2i}\right)^4\cdot\left(\frac{e^{2ix}+e^{-2ix}}2\right)^4$$
$$=\left(\frac{e^{4ix}-e^{-4ix}}{4i}\right)^4=\frac{e^{i16x}+e^{-i16x}-\binom41(e^{i8x}+e^{-i8x})+\binom42}{256\cdot i^4}=\frac{2\cos16x-4(2\cos8x)+6}{256}=\frac{\cos16x-4\cos8x+3}{128}$$

