integral of $\int\frac{2\sin(2x)-\cos(x)}{6-\cos^2x -4\sin x}\mathrm{d}x$ so i got this problem $$\int\dfrac{2\sin 2x-\cos x}{6-\cos^2x -4\sin x}\mathrm{d}x$$
now this is what i tried $=\int\dfrac{4\sin(x)\cos(x)-\cos(x)}{6-(1-\sin^2x) -4\sin x}\mathrm{d}x=\int\dfrac{\cos(x)(4\sin(x)-1)}{5+\sin^2x-4\sin x}\mathrm{d}x$ 
Substituting $\sin x=t$
implying $\int\dfrac{(4t-1)}{5+t^2-4t}\mathrm{d}t$=$\int\dfrac{(4t)}{5+t^2-4t}\mathrm{d}t+\int\dfrac{(-1)}{5+t^2-4t}\mathrm{d}t$ 
we may now solve second part by substuting for $\arctan( x)$ but what to do with $\int\dfrac{(4t)}{5+t^2-4t}\mathrm{d}t$ ??

$\large Hint:$ $u \equiv \sin\left(x\right)$.
\begin{align}
&
\int
{2\sin\left(2x\right) - \cos\left(x\right)
\over
6 - \cos^{2}\left(x\right) - 4\sin{x}}\,{\rm d}x
=
\int
{4\sin\left(x\right) - 1
\over
\sin^{2}\left(x\right) - 4\sin{x} + 5}\,\cos\left(x\right)\,{\rm d}x
\\[3mm]&=
\int{4u - 1 \over u^{2} - 4u + 5}\,{\rm d}u
=
\int{4\left(u - 2\right) + 7 \over \left(u - 2\right)^{2} + 1}\,{\rm d}u
=
2\ln\left(\left[u - 2\right]^{2} + 1\right)
+
7\arctan\left(u - 2\right)
\\[3mm]&=
2\ln\left(\sin^{2}\left(x\right) - 4\sin{x} + 5\right)
+
7\arctan\left(\sin\left(x\right) - 2\right)
\end{align}
$+$ a constant.
