Find $\int_{-5}^5 \sqrt{25-x^2}~dx$ I need to evaluate $$\int_{-5}^5 \sqrt{25-x^2}~dx$$
How would I do it? 
 A: Easy way is what @AndreNicolas said.  Calculus approach is trig substitution (here is a link)
So we let 
$y=\sqrt{25-x^2}$
Let x = $5\sin(\theta)$  , $dx = 5\cos(\theta)d\theta$
$y=\sqrt{25-(5\sin(\theta))^2}$=$\sqrt{25-25\sin^2(\theta)}$=$5\sqrt{1-\sin^2(\theta)}$=$5\cos(\theta)$
So your integral becomes $\int^{\sin^{-1}(1)}_{\sin^{-1}(-1)}25\cos^2(\theta)d\theta$.  When you do this out you get 25$\pi$/2.  Coincidence?
A: $$\int_{-5}^5 \sqrt {25 - x^2} \,dx = \text{area of semi-circle of radius 5}$$
That is, $y = \sqrt{25 - x^2},\;$ defines the top half of the circle $x^2 + y^2 = 5^2 = 25,\;$ which is centered at the origin, and has a radius of $5$. See the "real" portion of the graph below, outlined in blue:


If you are taking calculus, then we need for you to get comfortable with computing area using integrals! Short cuts won't help in the long run. Besides, it's fun, once you get the hang of it. 
So let's do Calculus!:  you can integrate the integral fairly easily by using trigonometric substitution. 
In this case, let's put $\,\bf x = 5\sin \theta$. Then $\bf\,dx = 5 \cos \theta\,d\theta.\,$ and $\theta = \sin^{-1}\left(\dfrac x5\right).$ Our bound of integration then become $(\theta = 3\pi/2 \;\text{to} \;\theta =\pi/2)$. But we'll simplify matters by doubling the area of the circle as theta sweeps from $0$ to $\pi/2$.
So, $$\begin{align}\int \sqrt{25 - x^2}\, dx & = \int \left(\sqrt{25 - 25\sin^2 \theta}\right)(5\cos \theta)\,d\theta \\ \\
& = 5\int \left(\sqrt{25(1 - sin^2 \theta}\right)\cos \theta\,d\theta\\ \\
& = 5\int 5\sqrt{\cos^2x} \cos \theta\,d\theta\\ \\
&= 25 \int \cos^2 \theta\,d\theta \\ \\
& = 25\cdot 2\int_0^{\pi/2}\left( \dfrac 12 + \dfrac {\cos 2\theta}2\right)\,d\theta\end{align}$$ 
In the last step we simply use the identity: $$\cos^2 \theta = \dfrac{1 + \cos 2 \theta}{2}$$. Then it's just a matter of using the power rule, and you're almost there.
A: If you are at the earliest stages of integration, you are probably supposed to do this problem  without using any integration techniques.
The curve $y=\sqrt{25-x^2}$ is the top half of a circle with centre the origin and radius $5$.
Our integral $\int_{-5}^5 \sqrt{25-x^2}\,dx$ is the area under the curve $y=\sqrt{25-x^2}$ and above the $x$ axis, from $x=-5$ to $x=5$.  That's a half-circle. Thus by a familiar formula the  area is $25\pi/2$.
