integration of trigonometry functions I have question regarding the following two integrals;
$\int_0^{2\pi} \frac{\sin\theta}{A-\sin\theta} d\theta$, and 
$\int_0^{2\pi} \frac{1}{A-\sin\theta} d\theta$ where $A>0$.
What method can be used to integrate them.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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$\ds{\int_{0}^{2\pi}{\sin\pars{\theta} \over A - \sin\pars{\theta}}\,\dd \theta =
-2\pi + A\int_{0}^{2\pi}{\dd\theta \over A - \sin\pars{\theta}}}$.
  I'll assume $\ds{A \in \mathbb{R}\setminus\bracks{-1,1}}$.

\begin{align}
\int_{0}^{2\pi}{\dd\theta \over A - \sin\pars{\theta}} & =
\int_{-\pi}^{\pi}{\dd\theta \over A + \sin\pars{\theta}} =
\int_{0}^{\pi}\bracks{{1 \over A + \sin\pars{\theta}} +
{1 \over A - \sin\pars{\theta}}}\dd\theta
\\[5mm] & =
2A\int_{0}^{\pi}{\dd\theta \over A^{2} - \sin^{2}\pars{\theta}} =
2A\int_{-\pi/2}^{\pi/2}{\dd\theta \over A^{2} - \cos^{2}\pars{\theta}} =
4A\int_{0}^{\pi/2}{\dd\theta \over A^{2} - \cos^{2}\pars{\theta}}
\\[5mm] & =
4A\int_{0}^{\pi/2}{\sec^{2}\pars{\theta} \over A^{2}\sec^{2}\pars{\theta} - 1}
\,\dd\theta =
4A\int_{0}^{\pi/2}{\sec^{2}\pars{\theta} \over
A^{2}\tan^{2}\pars{\theta} + A^{2} - 1}\,\dd\theta
\\[5mm] & =
4A\,{1 \over A^{2} - 1}\,{\root{A^{2} - 1} \over \verts{A}}\int_{0}^{\pi/2}
{\verts{A}\sec^{2}\pars{\theta}/\root{A^{2} - 1} \over
\bracks{\verts{A}\tan\pars{\theta}/\root{A^{2} - 1}}^{2} + 1}\,\dd\theta
\\[5mm] & =
{4\,\mrm{sgn}\pars{A} \over \root{A^{2} - 1}}\
\underbrace{\int_{0}^{\infty}{\dd t \over t^{2} + 1}}_{\ds{=\ {\pi \over 2}}}\ =\
\bbx{2\pi\,{\mrm{sgn}\pars{A} \over \root{A^{2} - 1}}}
\end{align}

where
  $\ds{t = {\verts{A} \over \root{A^{2} - 1}}\,\tan\pars{\theta}}$.

A: Hint: substitute $$\sin(\theta)=\frac{2t}{1+t^2}$$ and $$d\theta=\frac{2}{1+t^2}dt$$
Your new integral is given by $$\int 4\,{\frac {t}{ \left( A{t}^{2}+A-2\,t \right)  \left( {t}^{2}+1
 \right) }}
dt$$
to integrate use that $$\frac{4t}{(At^2-2t+A)(t^2+1)}=4+1/5\,{\frac {1}{ \left( t-2 \right) A}}+1/5\,{\frac {-t-2}{ \left( {
t}^{2}+1 \right) A}}
$$
A: Hint:
Consider the substitution
$$u=\sin \theta$$
and the identity
$$\cos\theta =\pm\sqrt{1-\sin^2\theta}$$
