<s>[INST] Does $\int_{0}^{\pi/ n}{\sin^3(x)\over [1+\cos^2(x)]^2}\mathrm dx=F(n)$ have a closed form? 
$$\int_{0}^{\pi/ n}{\sin^3(x)\over [1+\cos^2(x)]^2}\mathrm dx=F(n)\tag1$$

For $n\ge 1$
I have calculated out for:
$F(1)=1$
$F(2)={1\over 2}$
$F(3)={1\over 10}$
The general for $F(n)$ I wasn't able to find...

What is the closed form for $(1)?$

An attempt
Using $\cos^2(x)={1+\cos(2x)\over 2}$ then $1+\cos^2(x)={3+\cos(2x)\over 2}$
Calling upon a subsitition $u=3+\cos(2x)$ then $du=-2\sin(2x)dx$
Then $(1)$ becomes:
$$-\int_{1}^{\cos(2\pi/n)}{\sin^2(x)\over u^2\cos(x)}\mathrm du\tag2$$
$2+2\cos^2(x)=u$ then $\sin^2(x)={4-u\over 2}$
$\cos(x)=\sqrt{u-2\over 2}$
$${\sqrt{2}\over 2}\int_{1}^{\cos(\pi/n)}{u-4\over u^2\sqrt{u-2}}\mathrm du\tag3$$
$${\sqrt{2}\over 2}\int_{1}^{\cos{(\pi/n)}}{1\over u\sqrt{u-2}}\mathrm du+{2\sqrt{2}}\int_{1}^{\cos(\pi/n)}{1\over u^2\sqrt{u-2}}\mathrm du\tag4$$
From Standard integral table

$$\int{dx\over x\sqrt{ax+b}}={2\over \sqrt{-b}}\arctan\sqrt{ax+b\over -b}$$
$$\int{dx\over x^2\sqrt{ax+b}}=-{\sqrt{ax+b}\over bx}-{a\over 2b}\int{dx\over x\sqrt{ax+b}}$$

Substitute it in,
$${\sqrt{2}\over 2}\cdot{2\over\sqrt{2}}\arctan\sqrt{u-2\over 2}-2\sqrt{2}\left[-{\sqrt{u-2\over -2u}}+{1\over 4}\left({2\over\sqrt{2}}\cdot\arctan{\sqrt{u-2\over2}}\right)\right]\tag5$$
Break down to
$$\arctan\sqrt{u-2\over 2}-\sqrt{2}\cdot{\sqrt{u-2}\over u}-\arctan\sqrt{u-2\over 2}\tag6$$
$$=-\sqrt{2}\cdot{\sqrt{u-2}\over u}\tag7$$
I don't thing I am on the right track, any help?
 [/INST] Hint:
\begin{align*}
F(n)&=-\int_{0}^{\pi/ n}{\sin^2(x)\over [1+\cos^2(x)]^2}\mathrm d\cos x\\
&=-\int_{0}^{\pi/ n}{1-\cos^2(x)\over [1+\cos^2(x)]^2}\mathrm d\cos x\\\tag1
&=\int_{\cos\left ( \frac{\pi }{n} \right )}^{1}\frac{1-x^{2}}{\left ( 1+x^{2} \right )^{2}}\, \mathrm{d}x\\
&=\frac{1}{2}-\frac{\cos\left (  \dfrac{\pi }{n}\right )}{1+\cos^2\left (  \dfrac{\pi }{n}\right )}
\end{align*}
$(1):$ $\displaystyle \int \frac{1-x^{2}}{\left ( 1+x^{2} \right )^{2}}\, \mathrm{d}x=\frac{x}{1+x^{2}}+C$
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