# Math Help - Another seemingly simple integral

1. ## 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...

2. $\int_{-1/2}^{1/2}{\frac{9}{\sqrt{1-x^{2}}}\,dx}=2\int_{0}^{1/2}{\frac{9}{\sqrt{1-x^{2}}}\,dx}.$

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

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

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?

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

5. 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?
$\int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{9}{\sqrt{1-x^2}}dx=
\int_{-\frac{1}{2}}^{0}\frac{9}{\sqrt{1-x^2}}dx+\int_{0}^{\frac{1}{2}}\frac{9}{\sqrt{1-x^2}}dx$
,
but $\frac{9}{\sqrt{1-(-x)^2}}=\frac{9}{1-x^2}\Rightarrow \int_{-\frac{1}{2}}^{0}\frac{9}{\sqrt{1-x^2}}dx=\int_{0}^{\frac{1}{2}}\frac{9}{\sqrt{1-x^2}}dx$.
Therefore $\int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{9}{\sqrt{1-x^2}}dx=
2\int_{0}^{\frac{1}{2}}\frac{9}{\sqrt{1-x^2}}dx=18\int_{0}^{\frac{1}{2}}\frac{dx}{\sqrt{1-x^2}}$
.

PD: Late >.<

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