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

[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

 

[Hootenanny]

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

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.

 

[momogiri]

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

 

[Hootenanny]

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

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