How to show that $\int_{0}^{\pi}\sin^3(x){\mathrm dx\over (1+\cos^2(x))^2}=1?$ How can we show that $(1)$
$$\int_{0}^{\pi}\sin^3(x){\mathrm dx\over (1+\cos^2(x))^2}=1?\tag1$$

$\sin^3(x)={3\over 4}\sin(x)-{1\over 4}\sin(3x)$
$1+\cos^2(x)=2-\sin^2(x)$
$$\int_{0}^{\pi}[{3\over 4}\sin(x)-{1\over 4}\sin(3x)]{\mathrm dx\over (2-\sin^2(x))^2}=1\tag2$$
$2-\sin^2(x)={3\over 2}+{1\over 2}\cos(2x)$
$$\int_{0}^{\pi}{3\sin(x)-\sin(3x)\over 3+\cos(2x)}\mathrm dx\tag3$$
$$\int_{0}^{\pi}{3\sin(x)\over 3+\cos(2x)}\mathrm dx-\int_{0}^{\pi}{\sin(3x)\over 3+\cos(2x)}\mathrm dx\tag4$$
$\cos(2x)={1-\tan^2(x)\over 1+\tan^2(x)}$
$\sin(x)={2\tan(x/2)\over 1+\tan^2(x/2)}$
 A: Well, we have:
$$\mathscr{I}_{\space\text{n}}:=\int_0^\text{n}\frac{\sin^3\left(x\right)}{\left(1+\cos^2\left(x\right)\right)^2}\space\text{d}x\tag1$$
Substitute $\text{u}:=\cos\left(x\right)$:
$$\mathscr{I}_{\space\text{n}}:=\int_1^{\cos\left(\text{n}\right)}\frac{\text{u}^2-1}{\left(1+\text{u}^2\right)^2}\space\text{d}\text{u}=\int_1^{\cos\left(\text{n}\right)}\frac{1}{1+\text{u}^2}\space\text{d}\text{u}-2\int_1^{\cos\left(\text{n}\right)}\frac{1}{\left(1+\text{u}^2\right)^2}\space\text{d}\text{u}\tag2$$
Now, we know:
$$\int_1^{\cos\left(\text{n}\right)}\frac{1}{1+\text{u}^2}\space\text{d}\text{u}=\left[\arctan\left(\text{u}\right)\right]_1^{\cos\left(\text{n}\right)}=\arctan\left(\cos\left(\text{n}\right)\right)-\arctan\left(1\right)=$$
$$\arctan\left(\cos\left(\text{n}\right)\right)-\frac{\pi}{4}\tag3$$
We can write:
$$2\int_1^{\cos\left(\text{n}\right)}\frac{1}{\left(1+\text{u}^2\right)^2}\space\text{d}\text{u}=\left[\frac{\text{u}}{1+\text{u}^2}\right]_1^{\cos\left(\text{n}\right)}+\int_1^{\cos\left(\text{n}\right)}\frac{1}{1+\text{u}^2}\space\text{d}\text{u}=$$
$$\left[\frac{\text{u}}{1+\text{u}^2}\right]_1^{\cos\left(\text{n}\right)}+\left[\arctan\left(\text{u}\right)\right]_1^{\cos\left(\text{n}\right)}=$$
$$\frac{\cos\left(\text{n}\right)}{1+\cos^2\left(\text{n}\right)}-\frac{1}{1+1^2}+\arctan\left(\cos\left(\text{n}\right)\right)-\arctan\left(1\right)=$$
$$\frac{\cos\left(\text{n}\right)}{1+\cos^2\left(\text{n}\right)}-\frac{1}{2}+\arctan\left(\cos\left(\text{n}\right)\right)-\frac{\pi}{4}\tag4$$
So, we end up with:
$$\mathscr{I}_{\space\text{n}}=\arctan\left(\cos\left(\text{n}\right)\right)-\frac{\pi}{4}-\left\{\frac{\cos\left(\text{n}\right)}{1+\cos^2\left(\text{n}\right)}-\frac{1}{2}+\arctan\left(\cos\left(\text{n}\right)\right)-\frac{\pi}{4}\right\}=$$
$$\frac{1}{2}-\frac{\cos\left(\text{n}\right)}{1+\cos^2\left(\text{n}\right)}\tag5$$
A: Beside the hint MyGlasses gave, another simple way would be to directly use the tangent half-angle substitution which makes 
$$\int{\sin^3(x)\over (1+\cos^2(x))^2}\,dx=\int\frac{4 t^3}{\left(1+t^4\right)^2}\,dt=\int \frac{d(t^4)}{(1+t^4)^2}=\int \frac {du}{(1+u)^2}=-\frac 1 {1+u}$$
A: $$
\begin{align}
\int_0^\pi\sin^3(x)\frac{\mathrm{d}x}{(1+\cos^2(x))^2}
&=-\int_0^\pi\left(1-\cos^2(x)\right)\frac{\mathrm{d}\cos(x)}{\left(1+\cos^2(x)\right)^2}\\
&=\int_{-1}^1\left(1-u^2\right)\frac{\mathrm{d}u}{\left(1+u^2\right)^2}\\
&=\int_{-\pi/4}^{\pi/4}\left(1-\tan^2(v)\right)\frac{\mathrm{d}v}{\sec^2(v)}\\
&=\int_{-\pi/4}^{\pi/4}\left(\cos^2(v)-\sin^2(v)\right)\mathrm{d}v\\
&=\int_{-\pi/4}^{\pi/4}\cos(2v)\,\mathrm{d}v\\
&=\left[\frac12\sin(2v)\right]_{-\pi/4}^{\pi/4}\\[9pt]
&=1
\end{align}
$$
A: Hint:
With 
$$\int_0^\pi\dfrac{\sin^3x}{(1+\cos^2x)^2}dx=\int_0^\pi\dfrac{1-\cos^2x}{(1+\cos^2x)^2}\sin x\,dx$$
and let substitution $\cos x=u$.
A: For the integrand, use the trigonometric identitity $\sin^2(x)=1-\cos^2(x)$ and consider the two changes of variables \begin{align*}
u=\cos(x), &\,\,du=-\sin(x)\,dx\\
u=\tan(s), &\,\,du=\sec^2(s)\,ds
\end{align*}
to get\begin{align*}
\int_{0}^{\pi}\sin^3(x){\mathrm dx\over (1+\cos^2(x))^2}&=\int_0^\pi\frac{\sin(x)(1-\cos^2(x))}{(\cos^2(x)+1)^2}\,dx=-\int_{-1}^1\frac{1-u^2}{(u^2+1)^2}   \,du\\
&=-\int_{-1}^1\frac{1}{u^2+1}\,du+2\int_{-1}^1\frac{1}{(u^2+1)^2}\,du\\
&=\left[-\tan^{-1}(u)  \right]_{-1}^1+2\int_{-1}^1\frac{1}{(u^2+1)^2}\,du\\
&=-\frac{\pi}{2}+2\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac12\cos(2s)+\frac12\,ds\\
&=-\frac{\pi}{2}+1+\frac{\pi}{4}+\frac{\pi}{4}=1.
\end{align*}
