<s>[INST] $\int_{-1}^1\frac{y}{\pi\sqrt{1-y^2}}\,dy$ The solution says that we will get this $$\left[-\frac{\sqrt{1-y^2}}{\pi} \right]^1_{-1} $$. 
How do we get this definite integral. I know we are integrating an odd function over symmetric domain so the answer will be 0.
 [/INST] Let $u = 1-y^2$, then $du = -2y ~dy$ so that 
\begin{align}
\int_{-1}^{1} \frac{y}{\pi\sqrt{1-y^2}} dy &= \int_{y=-1}^{y=1} \frac{\frac{-1}{2}du }{\pi\sqrt{u}} = -\frac{1}{2\pi}\int_{y=-1}^{y=1} u^{\frac{-1}{2}}du = -\frac{1}{2\pi}\left(2u^{\frac{1}{2}}\right)_{y=-1}^{y=1}\\
&=-\left(\frac{\sqrt{1-y^2}}{\pi}\right)_{y=-1}^{y=1}
\end{align}
</s>