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...

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- Feb 25th 2009, 01:02 PMfattydqAnother seemingly simple integral
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... - Feb 25th 2009, 01:05 PMKrizalid
$\displaystyle \int_{-1/2}^{1/2}{\frac{9}{\sqrt{1-x^{2}}}\,dx}=2\int_{0}^{1/2}{\frac{9}{\sqrt{1-x^{2}}}\,dx}.$

Look at a table of integrals, that primitive is the arcsin function. - Feb 25th 2009, 01:10 PMfattydq
- Feb 25th 2009, 01:22 PMKrizalid
$\displaystyle \int_{-a}^af=2\int_0^af$ provided that $\displaystyle f$ is an even function and $\displaystyle a\in\mathbb R.$

- Feb 25th 2009, 01:26 PMAbu-Khalil
$\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 >.< - Feb 25th 2009, 01:44 PMfattydq