I became curious about the alternative methods to write a formula for the area of a circle rather than the popular method of dividing a circle into equal sections of an infinitesimal angle.
I gave it a thought and made a little headway.
For a circle, $x^2 + y^2 = R^2$, where $R$ is the radius.
$\therefore y = \sqrt{R^2 - x^2}$
$\int_{0}^{R} \sqrt{R^2 - x^2} dx$ = Area of each quadrant
$\Rightarrow$ Area of circle = $\int_{0}^{R} 4\sqrt{R^2 - x^2} dx$
I can't understand how to start with.
To compute this, we can use the standard trig-sub, let $x=R\sin\theta$, so $x\in[0, R]\Rightarrow\theta\in[0,\frac\pi2]$. Also, we have $dx=R\cos\theta d\theta$, so the integral is converted to
$$I=\int_{0}^{\frac\pi2} \sqrt{R^2 - R^2\sin^2\theta}\cdot R\cos\theta d\theta$$
We use the identity $1-\sin^2\theta=\cos^2\theta$, and get
$$I=\int_{0}^{\frac\pi2} R\cos\theta\cdot R\cos\theta d\theta=R^2\int_{0}^{\frac\pi2} \cos^2\theta ~d\theta$$
Next, we use identity $\cos^2\theta=\frac{1+\cos2\theta}2$
$$I=\frac{1}2R^2\int_{0}^{\frac\pi2} 1+\cos2\theta ~d\theta=\frac{1}2R^2\frac{\pi}{2}+\frac{1}2R^2\cdot\frac12\sin2\theta\bigg|_0^\frac\pi2$$
Note the second term vanishes, hence we get
$$I=\frac14\pi R^2$$
which is the area of a quarter of the circle.