# Thread: [SOLVED] Double check my work?

1. ## [SOLVED] Double check my work?

Determine the integral from 0 to 1 of 3/(sqrt(4-x^2)) dx

Here's what I did:

3/ sqrt(4(1-(x^2)/4))
(3/2)(1/(sqrt(1-(x/2)^2))
3/2 0 to 1 1/(sqrt(1-u^2)) where u=x/2
3/2 (sin^-1(x))] 0 to 1
(3/2)*(pi/2)=3pi/4

Did I do this correctly? If not, could someone please point out my mistakes? Thanks in advance!

2. Originally Posted by yzobel
Determine the integral from 0 to 1 of 3/(sqrt(4-x^2)) dx

Here's what I did:

3/ sqrt(4(1-(x^2)/4))
(3/2)(1/(sqrt(1-(x/2)^2))
3/2 0->1 1/(sqrt(1-u^2)) where u=x/2
3/2 (sin^-1(x))] 0->1
(3/2)*(pi/2)=3pi/4

Did I do this correctly? If not, could someone please point out my mistakes? Thanks in advance!
$\displaystyle \int{\frac{3}{\sqrt{4 - x^2}}\,dx}$.

Use a trigonometric substitution with $\displaystyle x = 2\sin{\theta}$ so that $\displaystyle dx = 2\cos{\theta}\,d\theta$.

Then the integral becomes

$\displaystyle \int{\frac{3}{\sqrt{4 - (2\sin{\theta})^2}}\,2\cos{\theta}\,d\theta}$

$\displaystyle = \int{\frac{6\cos{\theta}}{\sqrt{4 - 4\sin^2{\theta}}}\,d\theta}$

$\displaystyle = \int{\frac{6\cos{\theta}}{\sqrt{4(1 - \sin^2{\theta})}}\,d\theta}$

$\displaystyle = \int{\frac{6\cos{\theta}}{2\sqrt{\cos^2{\theta}}}\ ,d\theta}$

$\displaystyle = \int{\frac{3\cos{\theta}}{\cos{\theta}}\,d\theta}$

$\displaystyle = \int{3\,d\theta}$

$\displaystyle = 3\theta + C$.

Now remembering that $\displaystyle x = 2\sin{\theta}$

$\displaystyle \frac{x}{2} = \sin{\theta}$

$\displaystyle \theta = \arcsin{\frac{x}{2}}$.

Therefore, the integral is

$\displaystyle 3\theta + C = 3\arcsin{\frac{x}{2}} + C$.

So, putting in the limits...

$\displaystyle \int_0^1{\frac{3}{\sqrt{4 - x^2}}\,dx} = \left[3\arcsin{\frac{x}{2}}\right]_0^1$

$\displaystyle = 3\arcsin{\frac{1}{2}} - 3\arcsin{\frac{0}{2}}$

$\displaystyle = \frac{3\pi}{6} - 3(0)$

$\displaystyle = \frac{\pi}{2}$.