How to find $\int \sqrt{a^2-x^2} dx$, where $a$ is a constant How to find $$\int \sqrt{a^2-x^2} dx\;,$$ where $a$ is a constant?
It appears to be $$\frac{\pi a}{2}\;,$$ but how do I get there? 
 A: One way of finding the definite integral is to consider the shape of the graph.  If $y = \sqrt{a^2-x^2}$ then $y^2=a^2-x^2$, so $x^2+y^2=a^2$.  Apply the Pythagorean theorem: that's the equation of a circle of radius $a$.  $y = \pm\sqrt{a^2-x^2}$ is the whole circle; $y = \sqrt{a^2-x^2}$ is the top half of the circle.  If you know that the area of the whole circle is $\pi a^2$, then the area of the top half is
$$
\int_{-a}^a \sqrt{a^2-x^2} \, dx = \frac{\pi a^2}{2}.
$$
Generally, if you have
$$
a^2-x^2
$$
in an integral, you can use $x = a\sin\theta$ and $dx = a\cos\theta\,d\theta$, and then $a^2-x^2$ becomes $a^2\cos^2\theta$.
If you have
$$
a^2 + x^2
$$
then you can use $x = a\tan\theta$, $dx = a\sec^2\theta\,d\theta$, $a^2+x^2=a^2\sec^2\theta$.
If you have
$$
x^2 - a^2
$$
then you can use $x=a\sec\theta$, $dx=a\sec\theta\tan\theta\,d\theta$, $x^2 - a^2 = a^2\tan^2\theta$.
A: It might be enlightening for you to think the following.
The function to integrate is $\sqrt{a^2-x^2}=y$.
This gives that
$$y^2+x^2=a^2$$
I really hope you know this is the equation for a circle of radius $a$ centered at $(0,0)$. Since you're integrating from $-a$ to $a$, you're calculating the area of half a circle of radius $a$. This means that the integral
$$\int_{-a}^{a} \sqrt{a^2-x^2}dx $$
gives the area of half a circle of radius $a$, which is $\dfrac 1 2\pi a^2$.
