Deriving area of sector through calculus

The area of a sector of a circle is the area of the triangle plus an additional portion which is $\int_{r cos\theta}^r \sqrt{r^2 - x^2} dx$

In order to integrate this, a trig substitution is used, $x =rsin\theta, dx = rcos\theta$. But by making that substitution the integrating limits would change from $\frac{\pi}{4}$ to $\frac{\pi}{2}$ since $r = rsin\theta$ and $sin^{-1}(1) = \frac{\pi}{2}$ and for the lower limit we would have $cos\theta = sin\theta$, which $\theta = \frac{\pi}{4}$

But that doesn't make it any easier to solve for the area formula. What is the proper derivation of the area of a sector using calculus?

Answer

The angle $\theta$ is fixed, it is given to you.

When you are integrating $\sqrt{r^2-x^2}$ using a trig substitution, you must not use $\theta$, that's taken.

There are plenty of letters left, Greek if you like, let $x=\sin \phi$. Or maybe use $x=\sin t$. Then everything will work nicely.

Remark: This is a very time consuming way to find the area of a sector with angle $\theta$. For the area of the sector, if $\theta$ is given in radians, is$\dfrac{\theta}{2\pi}$ times the area of the circle.

That gives area $\dfrac{\theta}{2}r^2$.