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.
 A: Without loss of generality, let $0\le x < \frac{\pi}{2}$. Define
$$ f(x)=\sin x \ln \left(\frac{1 + \sin x}{1 - \sin x}\right). $$
Then $f(0)=0,f'(0)=0$ and
\begin{eqnarray}
f''(x)=\frac{1}{8\cos^2x}(4 \cos (2 x)-\sin x \ln (\frac{1+\sin
   x}{1-\sin x})-\sin (3 x) \ln(\frac{1+\sin x}{1-\sin
   x})+12).
\end{eqnarray}
We want to show that $f''(x)\ge 4$. Note
$$ f''(x)-4=-\frac{1}{2} \tan x \sec x \left(-4 \sin (x)+\ln \left(\frac{1+\sin
   x}{1-\sin x}\right)+\cos (2 x) \ln \left(\frac{1+\sin x}{1-\sin
   x}\right)\right). $$
Define
$$g(x)= 4 \sin x-\ln \left(\frac{1+\sin x}{1-\sin x}\right)-\cos (2 x) \ln
   \left(\frac{1+\sin x}{1-\sin x}\right). $$ 
Now we show that $g(x)\ge0$ for $0\le x<\pi/2$. In fact, easy calculation shows
$$ g'(x)=2\sin(2x)\ln \left(\frac{1+\sin x}{1-\sin x}\right)\ge0$$
for $0\le x<\pi/2$ and hence $g(x)$ is increasing. So $g(x)\ge 0$ and hence
$$ f(x)=f(0)+f'(0)+\frac{1}{2}f''(c)x^2=\frac{1}{2}f''(c)x^2\ge 2x^2 $$ for
some $c\in(0,\pi/2)$.
A: 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
$$
2x^2\le\sin(x)\log\left(\frac{1+\sin(x)}{1-\sin(x)}\right)\le2\tan^2(x)\tag{5}
$$

A: $$\int_0^x \cos t \:dt = \sin x$$
$$\int_0^x \frac{1}{\cos t} \:dt = \frac{1}{2}\ln \left(\frac{1 + \sin x}{1 - \sin x}\right)$$
Let
$$f(t) = \sqrt{\cos t}$$
$$g(t) = \frac{1}{\sqrt{\cos t}}$$
Cauchy–Schwarz 
$$\left(\int_0^x f(t)g(t) \:dt\right)^2 \leq \int_0^x f^2(t) \:dt \int_0^x g^2(t) \:dt$$
proves the result.
A: Hint $$\ln\left(\frac{1+\sin x}{1-\sin x}\right)\approx 2\left(\sin x+\frac{\sin^3 (x)}{3}+...\right)$$ or use $\ln(1+x)=x-x^2/2+x^3/3...$ and $\ln(1-x)=-(x+x^2/2+x^3/3...) $ as $|x|<1$
