Integrating the trigonometric function So the problem goes as follows:-
$$\int{{\cos^2x+\sin2x}\over{(2\cos x-\sin x)^2}}~dx$$
My attempt is as follows:-
\begin{align*}\int{{\cos^2x+\sin2x}\over{(2\cos x-\sin x)^2}}~dx&=\int{{\cos^2x+2\sin x\cos x}\over{(2\cos x-\sin x)^2}}~dx\\
&=
\int{{(\cos x)(\cos x+2\sin x)}\over{(2\cos x-\sin x)^2}}~dx
\end{align*}
Now i could see that $\cos x+2\sin x $ is $-1$ times derivative of the denominator:-
$$\cos x+2\sin x=(-1){{d(2\cos x-\sin x)}\over{dx}}$$
And to handle the $\cos x$ term in the numerator i tried the following:-
$$2\cos x=t+\sin x \implies \cos x={{t+\sin x}\over{2}}$$
But the $\sin x$ term in the numerator is annoying.   

Recall that $\int -{f'\over f^2} dx= {1\over f}+C$, setting $f(x)=2 \cos x-\sin x $ your integral is ${1\over 2 \cos x-\sin x}$
