Proof for area of a circle

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 $(0,0)$

Thus the equation of the circle is $y=\sqrt{r^2-x^2}$

To find the area, we will take the indefinite integral for now.

$\displaystyle \int \sqrt{r^2-x^2} \mathbb{d}x$

From here we substitute $x=r\sin(u)$. Now, we take the derivative of each side and we get

$\dfrac{dx}{du}=r\cos(u)$

$dx=r\cos(u)du$

Substituting for x and dx we have

$\displaystyle \int \sqrt{r^2-x^2}~ \mathbb{d}x=\int \sqrt{r^2-r^2\sin^2(u)} (r\cos (u)) ~du$

Focusing on the square root, we will use the identity $1-\sin^2(a)=\cos^2(a)$

$\sqrt{r^2(1-\sin^2(u))}=r\cos(u)$

Now we have

$\displaystyle \int r^2\cos^2(u)~\mathbb{d}u$

$r^2\displaystyle \int \cos^2(u)~\mathbb{d}u$

Now, to integrate $\cos^2(u)$ we must use $\cos^2(a)=\frac{\cos{2a}}{2}+\frac{1}{2}$ aka the half angle identity (Spent 15 mins trying u-sub on this step, then went to wolfram, glanced at first step to integrate $\cos^2(u)$ and slapped my self on the forehead). Now focusing on the cos part of the eq above

$r^2\displaystyle \int \frac{\cos{2u}}{2}+\frac{1}{2} ~du$

Splitting terms

$r^2\left(\displaystyle \int \frac{\cos{2u}}{2} ~du+\int \frac{1}{2} du\right)$

$r^2\left(\displaystyle \int \frac{\cos{2u}}{2}~du+\frac{u}{2}\right)$

Doing a w-sub for u

$w=2u$

$\frac{dw}{dx}=2$

$dw=2du$

Substituting back

$\displaystyle \int \frac{\cos(w)}{4}~dw+\frac{u}{2}$

$\frac{\sin(w)}{4}+\frac{u}{2}$

Re-substituting for w

$\frac{\sin(2u)}{4}+\frac{u}{2}$

Using double angle formula $\sin(2a)=2\sin(a)\cos(a)$

$\frac{\sin(2u)}{4}+\frac{u}{2}=\frac{\sin(u)\cos(u)}{2}+\frac{u}{2}$

Going back to our original eq

$r^2\displaystyle \int \cos^2(u)~\mathbb{d}u=r^2\left(\frac{\sin(u)\cos(u)}{2}+\frac{u}{2}\right)$

Now comes the slightly hard part, we must resubstitute (jeez, I've said "Resubstitute/substitute back" like 10 times already) $x=r\sin(u)$. Looking at the sin/cos half of our eq we have

$(r\sin(u))(r\cos(u))$

Using $\cos(a)=\sqrt{1-sin^2(a)}$

$\frac{1}{2}(r\sin(u))(r\sqrt{1-sin^2(u)})=\frac{1}{2}(r\sin(u))(\sqrt{r^2-r^2sin^2(u)})$

Substituting

$\frac{1}{2}(x)(\sqrt{r^2-x^2})$

We now have

$r^2\left(\frac{\sin(u)\cos(u)}{2}+\frac{u}{2}\right)=\frac{1}{2}(x)(\sqrt{r^2-x^2})+\frac{r^2u}{2}$

We can rearrange $x=r\sin(u)$ to $u=\sin^{-1}\left(\frac{x}{r}\right)$. Substituting (said it again)

$\frac{1}{2}x\sqrt{r^2-x^2}+\frac{r^2u}{2}=\frac{1}{2}x\sqrt{r^2-x^2}+\frac{r^2}{2}\sin^{-1}\left(\frac{x}{r}\right)$

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 $\frac{1}{2}x\sqrt{r^2-x^2}+\frac{r^2}{2}\tan^{-1}\left(\dfrac{x}{\sqrt{r^2-x^2}}\right)$

Let me explain, if we have an angle u, and $\sin(u)=\frac{x}{r}$ then r is the hypotenuse and x is the leg opposite of angle u. By Pythagorean theorem, the other leg is $\sqrt{r^2-x^2}$. This is the adjacent to angle u.

$\therefore \sin^{-1}\left(\dfrac{opposite}{hypotenuse}\right)=\tan^{-1}\left(\dfrac{opposite}{adjacent}\right)=\tan^{-1}\left(\dfrac{opposite}{\sqrt{hypotenuse^2-opposite^2}}\right)=u$

$\therefore \sin^{-1}\left(\dfrac{x}{r}\right)=\tan^{-1}\left(\dfrac{x}{y}\right)=\tan^{-1}\left(\dfrac{x}{\sqrt{r^2-x^2}}\right)=u$

skip here

Now, to find the area, we need the definite integral from $x=-r$ to $x=r$. Thus we have

$\displaystyle \int_{-r}^{r} \sqrt{r^2-x^2} \mathbb{d}x=\frac{1}{2}x\sqrt{r^2-x^2}+\frac{r^2}{2}\sin^{-1}\left(\frac{x}{r}\right)$

Equating

$\left(\frac{1}{2}(r)\sqrt{r^2-(r)^2}+\frac{r^2}{2}\sin^{-1}\left(\frac{(r)}{r}\right)\right)-\left(\frac{1}{2}(-r)\sqrt{r^2-(-r)^2}+\frac{r^2}{2}\sin^{-1}\left(\frac{(-r)}{r}\right)\right)$

Simplifying.

$\frac{r^2}{2}\sin^{-1}(1)-\frac{r^2}{2}\sin^{-1}(-1)=\frac{r^2}{2}\cdot\frac{\pi}{2}-\frac{r^2}{2}\cdot(-\frac{\pi}{2})$

$\Rightarrow \frac{r^2\pi}{2}$

Now, you may ask, "well isn't the formula $r^2\pi$ 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 $\pi r^2$

Fin.