$\int_{-1}^1\frac{y}{\pi\sqrt{1-y^2}}\,dy$ 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.
 A: $$I = \int_{-1}^{1} \frac{y}{\pi\sqrt{1-y^2}}dy = -\frac{1}{2}\int_{-1}^{1} \frac{\frac{d}{dy}(1-y^2)}{\pi\sqrt{1-y^2}}dy $$
$$I  = -\frac{1}{2}\left[(1-y^2)^\frac{1}{2}\cdot 2\cdot\frac{1}{\pi}\right]_{-1}^{1} =\left[-\frac{\sqrt{1-y^2}}{\pi}\right]_{-1}^{1}$$
A: The integrand is
$$\frac {-1}{\pi}\frac {-2y}{2\sqrt {1-y^2}} $$
but $$\int \frac {du}{2\sqrt {u}}=\sqrt {u} $$
observe and finish.
the final result is zero.
A: $$\int_{-1}^1\frac{y}{\pi\sqrt{1-y^2}}\,dy$$
Substitute $u=\sqrt{1-y^2}\implies du=-\frac{y}{\sqrt{1-y^2}}$. Then 
$$\int_{-1}^1\frac{y}{\pi\sqrt{1-y^2}}\,dy=-\int_{\text{limits}}\frac{du}{\pi}=-\left[\frac u\pi\right]_{\text{limits}}=\left[\frac{-\sqrt{1-y^2}}{\pi}\right]_{-1}^1$$ as required.
The motivation behind this substitution is to get rid of the awkward denominator (also by knowing with experience that the integrand is the exact derivative of this function, but I feel like this is a natural substitution to make even if you do not know this).
A: 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}
A: The integral equal to $0$ because the function is odd.
By the way, we have no any problem with substitution in primitive function  $y=\pm1$.
