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

[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

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Hootenanny
Staff Emeritus
Gold Member
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.

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

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