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

$$\int_{0}^{\pi}(1+2x)\cdot{\sin^3(x)\over 1+\cos^2(x)}\mathrm dx=(\pi+1)(\pi-2)\tag1$$

$(1)$ it a bit difficult to start with
$$\int_{0}^{\pi}(1+2x)\cdot{\sin(x)[1-\sin^2(x)]\over 1+\cos^2(x)}\mathrm dx\tag2$$
Setting $u=\cos(x)$ 
$du=-\sin(x)dx$
$$\int_{-1}^{1}(1+2x)\cdot{(u^2)\over 1+u^2}\mathrm du\tag3$$
$$\int_{-1}^{1}(1+2\arccos(u))\cdot{(u^2)\over 1+u^2}\mathrm du\tag4$$
$du=\sec^2(v)dv$
$$\int_{-\pi/4}^{\pi/4}(1+2\arccos(\tan(v)))\tan^2(v)\mathrm dv\tag5$$
$$\int_{-\pi/4}^{\pi/4}\tan^2(v)+2\tan^2(v)\arccos(\tan(v))\mathrm dv=I_1+I_2\tag6$$
$$I_1=\int_{-\pi/4}^{\pi/4}\tan^2(v)\mathrm dv=2-{\pi\over2}\tag7$$
As for $I_2$ I am sure how to do it.
 A: Using the fact that
$$
\int_0^\pi xf(\sin x)\,dx=\frac\pi2\int_0^\pi f(\sin x)\,dx
$$
and that $\cos^2=1-\sin^2x$ (so that this applies), you get that your integral equals
$$
(1+\pi)\int_0^\pi \frac{\sin^3 x}{1+\cos^2x}\,dx
$$
Writing
$$
\frac{\sin^3 x}{1+\cos^2x}=\frac{(1-\cos^2x)\sin x}{1+\cos^2x}=\frac{2}{1+\cos^2x}\sin x-\sin x
$$
we find that your integral equals
$$
(1+\pi)\bigl[-2\arctan(\cos x)+\cos x\bigr]_0^\pi=(1+\pi)(\pi-2).
$$
A: I am not sure this is the easiest (but it is a viable approach). Note that the antiderivative of $v'=\sin^3 x/(1+\cos^2 x)$ is given by $v=\cos x -2 \arctan \cos x$. Thus, we can integrate the problem by parts and obtain
$$I = (\cos x -2 \arctan \cos x)(1+2x)\Big|_{x=0}^\pi -2 \underbrace{\int_0^\pi(\cos x -2 \arctan \cos x)dx}_{=I_2}.$$
Now you can observe that $v=\cos x -2 \arctan \cos x$ is antisymmetric around $x=\pi/2$ (because $\cos$ is antisymmetric around $x=\pi/2$ and $\arctan$ is antisymmetric around $x=0$). Thus, the remaining integral vanishes $(I_2=0)$ and we obtain
$$I = (\cos x -2 \arctan \cos x)(1+2x)\Big|_{x=0}^\pi= (\pi+1)(\pi-2).$$
A: $J=\displaystyle \int_{0}^{\pi}(1+2x)\cdot{\sin^3(x)\over 1+\cos^2(x)}\mathrm dx$
Perform the change of variable $y=\pi-x$,
$\displaystyle J=\int_0^{\pi} (1+2(\pi-x))\dfrac{\sin^3 x}{1+\cos^2 x}dx$
Therefore,
$\begin{align}\displaystyle 2J&=(2+2\pi)\int_0^{\pi}\dfrac{\sin^3 x}{1+\cos^2 x}dx\\
2J&=-(2+2\pi)\int_0^{\pi}\dfrac{\sin^2 x}{1+\cos^2 x}\text{d}(\cos x)\\
2J&=-(2+2\pi)\int_0^{\pi}\dfrac{(1-\cos^2 x)}{1+\cos^2 x}\text{d}(\cos x)\\
\end{align}$
Perform the change of variable $y=\cos x$ in the latter integral,
$\begin{align}\displaystyle 2J&=(2+2\pi)\int_{-1}^{1}\dfrac{1-x^2}{1+x^2}dx\\
\displaystyle 2J&=(2+2\pi)\int_{-1}^{1}\dfrac{1}{1+x^2}dx-(2+2\pi)\int_{-1}^{1}\dfrac{x^2}{1+x^2}dx\\
&=(2+2\pi)\Big[\arctan x\Big]_{-1}^{1}-(2+2\pi)\left(\int_{-1}^1 \dfrac{1+x^2}{1+x^2}dx-\int_{-1}^1 \dfrac{1}{1+x^2}dx\right)\\
&=4(2+2\pi)\dfrac{\pi}{4}-2(2+2\pi)\\
&=2(\pi+1)(\pi-2)
\end{align}$
Therefore,
$\boxed{J=(\pi+1)(\pi-2)}$
A: HINT:
Use $\displaystyle I=\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$
so that $\displaystyle I+I=\int_a^b[f(x)+f(a+b-x)]dx$
Now use $\sin^3x\ dx=-(1-\cos^2x)d(\cos x)$
so replace $\cos x=u$
A: I am not sure if it is the correct approach,$$\int_{0}^{\pi}(1+2x)\cdot{\sin^3(x)\over 1+\cos^2(x)}\mathrm dx$$
Integrating by parts, taking $(1+2x)$ as first function and then we have:
$$(1+2x)\int_{0}^{\pi}\frac{sin^3x}{1+cos^2x}dx-2\int_{0}^{\pi}(\int_{0}^{v}\frac{sin^3x}{1+cos^2x}dx)dv$$
then for $\int_{0}^{\pi}\frac{sin^3x}{1+cos^2x}dx$
simplify it as: 
$$\int_{0}^{\pi}\frac{(1-cos^2x)sinx}{1+cos^2x}dx$$
substitute $cosx = t$
and there after it is fairly simple to calculate. by some method of integration.
