Prove: $\sin x \ln \left(\frac{1 + \sin x}{1 - \sin x}\right) \geq 2x^2$ 
Prove:  $$\sin x \ln \left(\frac{1 + \sin x}{1 - \sin x}\right) \geq 2x^2$$
  for $\: -\frac{\pi}{2} < x < \frac{\pi}{2}$


I have thought of three things so far that can be useful.
The Cauchy–Schwarz inequality 
$$
\int_Efg\,\mathrm{d}x\le\left(\int_Ef^2\,\mathrm{d}x\right)^{1/2}\left(\int_Eg^2\,\mathrm{d}x\right)^{1/2}
$$
Both sides of the inequality are even functions so its enough to consider 
$x \in [0, \frac{\pi}{2})$.
Maybe I can use that $\int_0^x \cos t \:dt = \sin x$.
I would appreciate a hint or a few on how to continue.
edit: The inequality is supposed to be proved using integration methods.

For $x^2\lt1$,
$$
\begin{align}
x\log\left(\frac{1+x}{1-x}\right)
&=\sum_{k=0}^\infty\frac{2x^{2k+2}}{2k+1}\\
&\le\sum_{k=0}^\infty2x^{2k+2}\\
&=\frac{2x^2}{1-x^2}\tag{1}
\end{align}
$$
If
$$
f(x)=\sin(x)\log\left(\frac{1+\sin(x)}{1-\sin(x)}\right)\tag{2}
$$
then
$$
f'(x)=\cos(x)\log\left(\frac{1+\sin(x)}{1-\sin(x)}\right)+2\tan(x)\tag{3}
$$
and applying $(1)$
$$
\begin{align}
f''(x)
&=2-\sin(x)\log\left(\frac{1+\sin(x)}{1-\sin(x)}\right)+2\sec^2(x)\\
&=4+2\tan^2(x)-\sin(x)\log\left(\frac{1+\sin(x)}{1-\sin(x)}\right)\\[4pt]
&\ge4+2\tan^2(x)-2\tan^2(x)\\[12pt]
&=4\tag{4}
\end{align}
$$
Since $f(0)=f'(0)=0$ and $f''(x)\ge4$, we get that $f(x)\ge2x^2$. Therefore, applying $(1)$ again, we get
$$
2x^2\le\sin(x)\log\left(\frac{1+\sin(x)}{1-\sin(x)}\right)\le2\tan^2(x)\tag{5}
$$

