Hi everyone. I'm going through a problem on definite integrals at the moment (attached picture). I went online to try and find some sort of guide for the solution and managed to find something close enough. It is on this link: https://www.quora.com/How-do-you-sol...sqrt-16-x-2-dx

When I first tried to solve the problem, I did the usual method of substituting u and du. That didn't seem to work... In the link above, I see that they substituted x with 4sint instead. Why is that the case? Can someone please explain to me the reason behind this?

Thank you very much for your responses!

The graph of $\displaystyle y = \sqrt{16-x^2}$ is the top half of the circle with radius 4 and centre the origin. What is the area under the curve from x=-4 to x=4 (without using calculus)?