Partial Fraction of $\int \frac{ \left( \cos x + \sin 2x \right) \ \mathrm{d}x}{( 2 - \cos^2 x)(\sin x)}$ If 
$$\int\frac{ \left( \cos x + \sin 2x \right) \ \mathrm{d}x}{( 2 - \cos^2 x)(\sin x)} = \int \frac{A \ \mathrm{d}x}{(\sin x)} + B \int\frac{\sin x \ \mathrm{d}x}{ 1 + \sin^2 x} + C \int \frac{\mathrm{d}x}{1 + \sin^2 x}$$
then, which of the following is correct?


*

*$A+B+C=4$

*$A+B+C=2$

*$A+BC=1$

*$A+B+C=5$
My approach is as follows: 
$$\int\frac{\left( \cos x + \sin 2x \right) \ \mathrm{d}x}{(2 - \cos^2 x)(\sin x)}= \int \frac{\left( \cos x + \sin 2x \right) \ \mathrm{d}x}{(1 + \sin^2 x)(\sin x)} =$$
$$= \int \frac{\cos x \ \mathrm{d}x}{( 1 + \sin^2 x)(\sin x)} + \int \frac{\sin 2x \ \mathrm{d}x}{(1 + \sin^2 x )(\sin x)} =$$
$$= \int \frac{ \cos x \ \mathrm{d}x}{(1 + \sin^2 x )(\sin x)} + \int \frac{ 2 \cos x \ \mathrm{d}x}{(1 + \sin^2 x )} =$$
$$= \int\frac{- \sin \cos x \ \mathrm{d}x}{ 1 + \sin^2 x } + \int\frac{\cos x \ \mathrm{d}x}{ \sin x } + \int \frac{2 \cos x \ \mathrm{d}x}{1 + \sin^2 x}$$
After this step I am not able to approach
 A: The integrand can be written
$$
\frac{\cos x+\sin2x}{(1+\sin^2x)\sin x}
$$
Let's try and determine $A$, $B$ and $C$ so that
$$
\frac{A}{\sin x}+\frac{B\sin x}{1+\sin^2x}+\frac{C}{1+\sin^2x}
$$
is the same. Removing the denominators this forces
$$
A(1+\sin^2x)+B\sin^2x+C\sin x=\cos x+\sin2x
$$
If we set $x=0$, we get $A=1$; with $x=\pi/2$, we get $2A+B+C=0$; with $x=-\pi/2$ we get $2A+B-C=0$.
The linear system
\begin{cases}
A=1\\
2A+B+C=0\\
2A+B-C=0
\end{cases}
has the solution $A=1$, $B=-2$, $C=0$.
Hence $A+BC=1$. However, the two functions are not equal as you can check at $\pi/4$.

In order to compute the integral, one has rather to find $A$, $B$ and $C$ such that
$$
\frac{1+2\sin x}{(1+\sin^2x)\sin x}=
\frac{A}{\sin x}+\frac{B\sin x}{1+\sin^2x}+\frac{C}{1+\sin^2x}
$$
which is possible and equivalent to solving for partial fractions
$$
\frac{1+2u}{u(1+u^2)}=\frac{A}{u}+\frac{Bu+C}{1+u^2}
$$
This certainly has a solution and allows to compute your integral with $u=\sin x$.
This translates into $A+Au^2+Bu^2+Cu=1+2u$, so
\begin{cases}
A+B=0\\
C=2\\
A=1
\end{cases}
so $A=1$, $B=-1$ and $C=2$.
With this fix, the right answer would be $A+B+C=2$.
A: Write your Integrand in the form
$$\frac {\csc(x)(\cos(x)+2\sin(x)\cos(x))}{2-\cos^2(x)}$$
and then multiply numerator and denominator by
$$\frac{csc(x)(\cos(x)+2\sin(x)\cos(x))}{2-\cos^2(x)}$$
and we get
$$\newcommand{\dx}{\; \mathrm{d}}x\int \frac{\cot(x)(\csc(x)+2)\csc(x)}{\csc^2(x)+1}\dx$$
now Substitute
$$u=\csc(x)$$
then
$$du=-\cot(x)\csc(x)dx$$
and we obtain
-$$\int\frac{u+2}{u^2+1}du$$
Can you finish?
