Answer
See explanations.
Work Step by Step
a. Given $f$ as an odd function, we have $f(-x)=-f(x)$. Letting $u=-x$, we have $du=-dx$ and the endpoints become $x=-a\to u=a$ and $x=a\to u=-a$.
We have
$F=\int_{-a}^af(x)dx=\int_a^{-a}f(-u)d(-u)=\int_a^{-a}f(u)du=-\int_{-a}^af(u)du=-F$
which gives $F=0$
b. Given $f(x)=sin(x)$, we have $f(-x)=sin(-x)=-sin(x)=-f(x)$.
Thus $f$ is an odd function.
We have $F=\int_{-\pi/2}^{\pi/2}f(x)dx=\int_{-\pi/2}^{\pi/2}sin(x)dx=-cos(x)|_{-\pi/2}^{\pi/2}=-cos(\pi/2)+cos(-\pi/2)=0$
which confirms the result from part (a).
