Integrate $ \sin x /(1 + A \sin x)$ over the range $0$,$2 \pi$ for $A=0.2$ Wolfram Alpha indicates the following solution form:-
$$
\int_0^{2\pi} \frac{\sin x}{1 + A \sin x} dx
= 
(1/A)\left( x - \frac{2 \tan^{-1} \left( \frac{A + \tan{(x/2)}}{\sqrt{(1-A^2)}}\right)}{\sqrt{(1-A^2)}} + constant\right)^{2\pi}_0
$$
My thinking is that substituting for $x$ by $2 \pi$ and by $0$ the corresponding values of $\tan(x/2)$ are the same, namely $\tan(\pi) = \tan(0) = 0$. It follows therefore that the fractional term:-
$$
\frac{2 \tan^{-1} \left( \frac{A + \tan{(x/2)}}{\sqrt{(1-A^2)}}\right)}{\sqrt{(1-A^2)}}
$$ 
has the same value ( let us call it $Q$ ) for $x=2\pi$ and $x=0.$ Therefore the definite integral becomes
$$
\int_0^{2\pi} \frac{\sin x}{1 + A \sin x} dx
= 
(1/A)\left( (2\pi - Q + constant) - (0 - Q + constant) \right)
= 2\pi/A.
$$
For $A = 0.2$ we derive the result 
$$
\int_0^{2\pi} \frac{\sin x}{1 + 0.2 \sin x} dx
= 10 \pi.
$$
However Wolfram Alpha gives the answer as $-0.64782$ (approx.) which is reasonable looking at the graph.
So what is wrong with my reasoning?  ( I realize that the $\tan$ function is not continuous over the range $0,2\pi$ but I don't know what that implies for the analysis of the definite integral ).
EDIT
I have accepted @abel's alternative method for solving the integral.
I have posted a related question here which asks why the approach which I followed is incorrect.
 A: $$\int_0^{2\pi}\frac{\sin x}{1+a\sin x }\, dx = \int_0^{2\pi}\left(\frac 1 a-\frac{1}{a(1+a\sin x)}\right)\, dx=\frac{2\pi}a-\frac1a \int_0^{2\pi}\frac{1}{1+a\sin x }\, dx =\frac{2\pi}a-\frac Ja $$
now consider 
$$\begin{align}J &= \int_0^{2\pi}\frac{1}{1+a\sin x }\, dx\\
&=\int_0^{\pi}\frac{1}{1+a\sin x }\, dx + \int_\pi^{2\pi}\frac{1}{1+a\sin x }\, dx \\
&=\int_0^{\pi}\frac{1}{1+a\sin x }\, dx + \int_0^\pi\frac{1}{1-a\sin x }\, dx \\
&=2\int_0^{\pi/2}\frac{1}{1+a\sin x }\, dx + 2\int_0^{\pi/2}\frac{1}{1-a\sin x }\, dx \\
&= 2f(a) + 2f(-a)\tag 1\end{align}$$
where $$f(a) = \int_0^{\pi/2}\frac{1}{1+a\sin x }\, dx,\quad  u = \tan(x/2), x = 2\tan^{-1}u, dx = \frac{2du}{1+u^2} $$ therefore 
$$\begin{align}f(a) &=2\int_0^1\frac{du}{1+2au+u^2} \\
&= 2\int_0^1\frac{du}{(u+a)^2 + 1- a^2}\\&=\frac{2}{\sqrt{1-a^2}} \tan^{-1}\left(\frac{u+a}{\sqrt{1-a^2}}\right)\big|_0^1\\
&=\frac{2}{\sqrt{1-a^2}}\left(\tan^{-1}\left(\frac{1+a}{\sqrt{1-a^2}}\right) - \tan^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right)\right)\\
&=\frac{2}{\sqrt{1-a^2}}\left(\tan^{-1}\left(\sqrt{\frac{1+a}{1-a}}\right) - \tan^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right)\right)\\
f(a) + f(-a) &= \frac{2}{\sqrt{1-a^2}}\left(\tan^{-1}\left(\sqrt{\frac{1+a}{1-a}}\right) + \tan^{-1}\left(\sqrt{\frac{1-a}{1+a}}\right)\right)\\ 
&=\frac{\pi}{\sqrt{1-a^2}}\end{align}$$
finally, $$\int_0^{2\pi}\frac{\sin x}{1+a\sin x }\, dx = \frac{2\pi}a-\frac{2\pi}{a\sqrt{1-a^2}}=\frac{2\pi(\sqrt{1-a^2} - 1)}{a\sqrt{1-a^2}}$$
