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

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.

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

Staff Emeritus