
Originally Posted by
zorro
$\displaystyle I_1$ = $\displaystyle 3 \left[ tan^{-1} (1) - tan^{-1} (-1) \right] + c_1$
There is no need to add a constant since $\displaystyle \int_{-1}^1\frac{1}{1+x^2}\,\mathrm{d}x$ is a definite integral.

Originally Posted by
zorro
= $\displaystyle \left[ \frac{6 \pi}{4} \right] + c_1$
$\displaystyle \frac{6\pi}{4}=\frac{3\times 2\pi}{2\times 2}=\frac{3\pi}{2}$

Originally Posted by
zorro
$\displaystyle I_2$ = $\displaystyle \int_{-1}^{1} \frac{sin^3 x}{1 + x^2} $ $\displaystyle dx_2$
$\displaystyle I_2 = \int_{-1}^0\frac{\sin^3x}{1+x^2}\,\mathrm{d}x + \int_0^1\frac{\sin^3x}{1+x^2}\,\mathrm{d}x$
Substitute $\displaystyle x=-t$ ($\displaystyle \implies t=-x$ and $\displaystyle \mathrm{d}t=-\mathrm{d}x$) in the first integral :
$\displaystyle \int_{-1}^0\frac{\sin^3x}{1+x^2}\,\mathrm{d}x =-\int_{1}^0\frac{\sin^3(-t)}{1+(-t)^2}\,\mathrm{d}t=\int_0^1\frac{-\sin^3t}{1+t^2}\,\mathrm{d}t=-\int_0^1\frac{\sin^3t}{1+t^2}\,\mathrm{d}t$
Thus $\displaystyle I_2 = -\int_0^1\frac{\sin^3t}{1+t^2}\,\mathrm{d}t+\int_0^ 1\frac{\sin^3x}{1+x^2}\,\mathrm{d}x=0.$
This is a special case of a more general result: Let $\displaystyle f:\mathbb{R}\to\mathbb{R}$ be an odd function and let $\displaystyle A$ be a real number. Then $\displaystyle \int_{-A}^Af(t)\,\mathrm{d}t=0.$
Edit : 
Originally Posted by
dedust
so $\displaystyle \int_{-1}^{1} \frac{\sin^3 x}{1 + x^2} \ \mathrm{d}x
= 0 $
Yes !