Finding value of $\displaystyle \int \frac{\cos(2x)\sin(4x)}{\cos^4(x)(1+\cos^2 (2x))}dx$
Try: let $$I =\int\frac{8\sin(2x)\cdot \cos^2(2x)}{(1+\cos 2x)^2\cdot (1+\cos^2(2x))}dx$$
Now put $\cos (2x)=t$ and $2\sin(2x)dx=-dt$
So $$I=-\int\frac{4t^2}{(1+t)^2\cdot (1+t^2)}dt$$
Could some help me to solve it without partial fraction. Thanks
A more useful substitution is $u=\tan^2 x$. (I started from the Weierstrass substitution, viz. $\tan x$ rather than $\tan\frac{x}{2}$ because the integrand has period $\pi$, but some experimenting showed $\tan^2 x$ makes it tidier.) Note that $\cos 2x=\frac{1-u}{1+u}$ and $du=\sin 2x\sec^4 xdx$. The calculations that follow obtain an answer which, based on Wolfram Alpha experiments, appear to be wrong: $$I=\int\frac{2\cos^2 2xdu}{(1+\cos^2 2x)^2}=\frac12\int\left(\frac{1-u^2}{1+u^2}\right)^2du=\frac12\int\left(\frac{2}{1+u^2}-1\right)^2du\\=\int\left(\frac{2}{(1+u^2)^2}-\frac{2}{1+u^2}+\frac12\right)du=\frac{\tan^2 x}{1+\tan^4 x}-\arctan(\tan^2 x)+\frac{\tan^2 x}{2}+C,$$where we've used $$\int\frac{2du}{(1+u^2)^2}=\frac{u}{1+u^2}+\arctan u+C$$(which you can obtain e.g. with $u=\tan\phi$).
I'm still trying to pinpoint errors, but this should be the best substitution to avoid partial fractions.