Easier way to solve $\int\frac{2+\sin x}{\sin x(1+\cos x)}dx$ Is there easier way than universal supstitution to solve this integral $$\int\frac{2+\sin x}{\sin x(1+\cos x)}dx$$?
 A: The integral $\int\frac{2}{\sin x(1+\cos x)}\,dx$ can be found by multiplying top and bottom by $\sin x$. After replacing the $\sin^2 x$ in the denominator by $1-\cos ^2 x$, we make the substitution $1-\cos x=u$. We end up with an integration that can be done using partial fractions.
For the remaining integral $\int \frac{1}{1+\cos x}\,dx$, use the fact this is $\int\frac{1}{2\cos^2(x/2)}\,dx$, so we want to integrate $\frac{1}{2}\sec^2 (x/2)$, easy.
A: First noting that $\frac{1+\cos(x)}{2}=\cos^2(x/2)$ and then using partial fractions,
$$
\begin{align}
&\int\frac{2+\sin(x)}{\sin(x)(1+\cos(x))}\mathrm{d}x\\
&=\int\frac{2\,\mathrm{d}x}{\sin(x)(1+\cos(x))}+\int\frac{\mathrm{d}x}{1+\cos(x)}\\
&=\int\frac{-2\,\mathrm{d}\cos(x)}{(1-\cos^2(x))(1+\cos(x))}+\int\sec^2(x/2)\,\mathrm{d}x/2\\
&=-2\int\left(\frac{1/4}{1-\cos(x)}+\frac{1/4}{1+\cos(x)}+\frac{1/2}{(1+\cos(x))^2}\right)\mathrm{d}\cos(x)+\tan(x/2)\\
&=\frac12\log\left(\frac{1-\cos(x)}{1+\cos(x)}\right)+\frac1{1+\cos(x)}+\frac{\sin(x)}{1+\cos(x)}+C\\
&=\log\left(\frac{\sin(x)}{1+\cos(x)}\right)+\frac{1+\sin(x)}{1+\cos(x)}+C
\end{align}
$$
Disclaimer: Having finally read André's answer, I see that my answer closely follows his outline. So consider this an implementation of his program.
A: We can rewrite the integral as $\displaystyle\int\frac{2}{\sin x(1+\cos x)} dx +\int \frac{1}{1+\cos x} dx$.
In the first integral, multiplying by $\sin x$ on the top and bottom gives $\displaystyle\int\frac {2\sin x}{(1-\cos^{2}x)(1+\cos x)} dx = \int\frac{-2}{(1+u)^2(1-u)} du$ after substituting $u=\cos x$,
and then partial fractions can be used to get 
$\displaystyle -\int\bigg(\frac{1/2}{1+u}+\frac{1}{(1+u)^2}+\frac{1/2}{1-u}\bigg) du=-\bigg(\frac{1}{2}\ln(1+u)-(1+u)^{-1}-\frac{1}{2}\ln(1-u)\bigg) du$
$\displaystyle=-\frac{1}{2}\ln(1+\cos x)+(1+\cos x)^{-1}+\frac{1}{2}\ln(1-\cos x)+C$
$\displaystyle =\ln\vert\csc x-\cot x\vert+(1+\cos x)^{-1}+C$.
In the second integral, multiplying by $1-\cos x$ on the top and bottom gives
$\displaystyle\int\frac{1-\cos x}{\sin^{2}x} dx=\int(\csc^{2}x-\csc x\cot x)\; dx = -\cot x+\csc x+C$.
A: Well, first step I see would be to multiply numerator and denominator by $1-\cos x$.
$$\int\dfrac{(1-\cos x)(2+\sin x)dx}{\sin^3x}=\int2\csc^3x+\csc^2x-2\csc^2 x\cot x-\csc x\cot xdx$$
The last 3 terms integrate quickly and easily and I'm guessing you've already seen how to handle $\int\csc^3xdx$.
