# Integral of {sqrt(4-x^2)} from 0 to 2?

1. Apr 23, 2008

### momogiri

[SOLVED] Integral of {sqrt(4-x^2)} from 0 to 2?

Can someone please explain step by step for this?
I know the answer is pi, I just don't understand the steps towards solving it

First, I'm supposed to set x = 2sin(u)
so then u = arcsin(x/2)
then somehow, {sqrt(4-x^2)} = 2cos(u)
How'd that happen? @__@
I'm really really confused!

Thanks for tolerating XD

2. Apr 23, 2008

### Hootenanny

Staff Emeritus
Okay so we have,

$$\int^{2}_{0} \sqrt{4-x^2}\;\;dx$$

And if we let $x = 2\sin u$, then we obtain,

$$\int^{2}_{0} \sqrt{4-\left(2\sin u\right)^2}\;\;dx$$

$$= \int^{2}_{0} \sqrt{4-4\sin^2 u}\;\;dx$$

$$= \int^{2}_{0} \sqrt{4\left(1-\sin^2 u\right)}\;\;dx$$

$$= \int^{2}_{0} 2\sqrt{1-\sin^2 u}\;\;dx$$

However, we know that $\sin^2\theta+\cos^2\theta = 1 \Rightarrow 1-\sin^2\theta = \cos^2\theta$. Hence, the integral becomes,

$$\int^{2}_{0} 2\cos u \;\;dx$$

Is that clear? Don't forget that before you can integrate, you need to make a change of variable from dx to du.

P.S. We have Homework & Coursework forums for textbook questions and assistance with homework and we have Mathematics forums for general mathematics discussions.

3. Apr 23, 2008

### momogiri

Ah! Yes! It helps very much!
Thank you!!
And sorry for the misplaced topic >__<;;

4. Apr 23, 2008

### Hootenanny

Staff Emeritus
A pleasure. And don't worry about misplacing your topic, we'll let you off... this time