So I when I woke up today, I decided to learn U-sub and trig-sub integration. About 30 mins after I started, I set myself to proving the area of a circle by integration. Here's what I got.
WLOG, assume the circle's center is located at
Thus the equation of the circle is
To find the area, we will take the indefinite integral for now.
From here we substitute . Now, we take the derivative of each side and we get
Substituting for x and dx we have
Focusing on the square root, we will use the identity
Now we have
Now, to integrate we must use aka the half angle identity (Spent 15 mins trying u-sub on this step, then went to wolfram, glanced at first step to integrate and slapped my self on the forehead). Now focusing on the cos part of the eq above
Doing a w-sub for u
Re-substituting for w
Using double angle formula
Going back to our original eq
Now comes the slightly hard part, we must resubstitute (jeez, I've said "Resubstitute/substitute back" like 10 times already) . Looking at the sin/cos half of our eq we have
We now have
We can rearrange to . Substituting (said it again)
You can skip down to the part that says "skip to here" if you don't want to read the next part, I just give an explanation of why wolfram's equation is slightly different than this one.
Now we are done. If you plug it into wolfram, they change the above equation slightly to
Let me explain, if we have an angle u, and then r is the hypotenuse and x is the leg opposite of angle u. By Pythagorean theorem, the other leg is . This is the adjacent to angle u.
Now, to find the area, we need the definite integral from to . Thus we have
Now, you may ask, "well isn't the formula what's up with the two in the denominator?"
Well, the graph of this equation is a semi-circle, not a full circle. So when we multiply our equation by two since we have the area of half the circle, we finally get our desired result of