<s>[INST] this integral $\int\frac{2-\sin{x}}{\sin^2{x}-\sin{x}+1}dx$ Find integral
$$\int\dfrac{2-\sin{x}}{\sin^2{x}-\sin{x}+1}dx$$
As suggested ,we take the Weierstrass Substitution is often useful in integrals such as the original.
$$
\begin{align}
\sin(x)&=\frac{2z}{1+z^2}\\
\cos(x)&=\frac{1-z^2}{1+z^2}\\
\mathrm{d}x&=\frac{2\,\mathrm{d}z}{1+z^2}
\end{align}
$$
I known is equal to
$$\int\dfrac{4(z^2-z+1)}{z^4-2z^3+6z^2-2z+1}dz$$
but I don't know how I should calculate the last integral.
 [/INST] Here's my revised solution that is shorter in working (which is still ridiculously long). The only tedious part is the simplifiying of the integrand after the Weierstrass Substitution. Outline:
Let $ y = \frac \pi 2 - x $, we get 
$ \begin{eqnarray} \displaystyle \int \frac{\cos y - 2}{\cos^2 y - \cos y + 1} \, dy & = & \displaystyle \int \frac{(-\sin y) + (\sin y + \cos y - 2)}{\cos^2 y - \cos y + 1} \, dy \\ & = & \int \frac{-\sin y}{\cos^2 y - \cos y + 1} \, dy + \int \frac{\sin y + \cos y - 2} {\cos^2 y - \cos y + 1} \, dy \end{eqnarray} $
The first integral can be solved via substitution of $ p = -\cos y $ then convert it in the form of $ \large\displaystyle \int \frac{ds}{a^2+s^2} = \frac1a \tan^{-1} \left( \frac sa\right) $
For the second integral, we use the  Weierstrass Substitution as suggested. Let $ t = \tan\left( \frac y2\right) $ and the integrand simplifies to
$ \large \frac{-6t^2+4t-2}{(t^2+1)^2} $
Apply partial fraction
$ \large \frac{4t}{(t^2+1)^2} - \frac6{t^2+1} + \frac4{(t^2+1)^2} $
Which can be easily integrated.
If I'm correct, back substituting everything yields
$ \large \frac2{\sqrt3} \tan^{-1} \left(\frac2{\sqrt3}\sin x\right) - \frac{2(\sin x + 1)}{\tan \frac x2 + 1} - 4\cot^{-1} x + C $
</s>