Evaluate $\displaystyle \int_{-8}^8\sqrt{64-x^2}\,dx$
the easy way: you could realize that this integral gives you the area of the upper half of the circle with radius 8
the hard way: use the trig substitution $\displaystyle x = 8 \sin \theta$ ....since we have an even function here, you may also wish to change the problem to $\displaystyle 2 \int_0^8 \sqrt{64 - x^2}~dx$, and change the limits based on the trig substitution