The definite integral from -1/2 to 1/2 of 9/the square root of 1-x^2 dx
I don't even know where to begin in this one. Finding the antiderivative of that seems impossible...
$\displaystyle \int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{9}{\sqrt{1-x^2}}dx=
\int_{-\frac{1}{2}}^{0}\frac{9}{\sqrt{1-x^2}}dx+\int_{0}^{\frac{1}{2}}\frac{9}{\sqrt{1-x^2}}dx$,
but $\displaystyle \frac{9}{\sqrt{1-(-x)^2}}=\frac{9}{1-x^2}\Rightarrow \int_{-\frac{1}{2}}^{0}\frac{9}{\sqrt{1-x^2}}dx=\int_{0}^{\frac{1}{2}}\frac{9}{\sqrt{1-x^2}}dx$.
Therefore $\displaystyle \int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{9}{\sqrt{1-x^2}}dx=
2\int_{0}^{\frac{1}{2}}\frac{9}{\sqrt{1-x^2}}dx=18\int_{0}^{\frac{1}{2}}\frac{dx}{\sqrt{1-x^2}}$.
PD: Late >.<