# Another seemingly simple integral

• Feb 25th 2009, 01:02 PM
fattydq
Another 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 PM
Krizalid
$\displaystyle \int_{-1/2}^{1/2}{\frac{9}{\sqrt{1-x^{2}}}\,dx}=2\int_{0}^{1/2}{\frac{9}{\sqrt{1-x^{2}}}\,dx}.$

Look at a table of integrals, that primitive is the arcsin function.
• Feb 25th 2009, 01:10 PM
fattydq
Quote:

Originally Posted by Krizalid
$\displaystyle \int_{-1/2}^{1/2}{\frac{9}{\sqrt{1-x^{2}}}\,dx}=2\int_{0}^{1/2}{\frac{9}{\sqrt{1-x^{2}}}\,dx}.$

Look at a table of integrals, that primitive is the arcsin function.

I know that, but what did you just do knowing that? Where did the 2 come from and why did the -1/2 just disappear? And I know that 1/the sq root of 1-x^2 but it's a 9 not a 1, so where does this come into play?
• Feb 25th 2009, 01:22 PM
Krizalid
$\displaystyle \int_{-a}^af=2\int_0^af$ provided that $\displaystyle f$ is an even function and $\displaystyle a\in\mathbb R.$
• Feb 25th 2009, 01:26 PM
Abu-Khalil
Quote:

Originally Posted by fattydq
I know that, but what did you just do knowing that? Where did the 2 come from and why did the -1/2 just disappear? And I know that 1/the sq root of 1-x^2 but it's a 9 not a 1, so where does this come into play?

$\displaystyle \int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{9}{\sqrt{1-x^2}}dx= \int_{-\frac{1}{2}}^{0}\frac{9}{\sqrt{1-x^2}}dx+\int_{0}^{\frac{1}{2}}\frac{9}{\sqrt{1-x^2}}dx$,
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 PM
fattydq
Quote:

Originally Posted by Krizalid
$\displaystyle \int_{-a}^af=2\int_0^af$ provided that $\displaystyle f$ is an even function and $\displaystyle a\in\mathbb R.$

I love calculus, because there's rule's for everything that I'm just supposed to know magically, intuitively. I swear my professor never mentioned anything like that.