integral of sqrt(4-x^2)dx
I've gathered that x = 4sin(theta) so I write it out as:
sqrt(4 - 2sin(theta))2cos(theta)
that's supposed to equal 4cos^2(theta) but I am getting stuck in the middle here
The reason we restrict to the interval is so that there is a one-to-one correspondence between the values of and the values of
Also, be sure to notice that Abu-Khalil's solution is not quite correct: if you substitute then the radicand becomes However, I'm sure he only made the error because he was carrying over the faulty substitution from your original post.