<s>[INST] Area of a circle I've tried to find as a personnal exercise where the formula $A=\pi R^2$ comes from.
After drawing the problem, I've found that $A = 2\int\limits_{-R}^{R}\sqrt{R^2-t^2}dt$. How can I calculate this ?
 [/INST] This is a classic case of trigonometric substitution. Set $t=R\sin\theta$, $-\frac{\pi}{2}\leq \theta\leq\frac{\pi}{2}$. 
Then $R^2 - t^2 = R^2(1-\sin^2\theta) = R^2\cos^2\theta$, hence $\sqrt{R^2-t^2} = \sqrt{R^2\cos^2\theta} = |R\cos\theta| = R\cos\theta$, because $R\geq 0$ and $\cos\theta\geq 0$ for $\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]$.
Since $t=R\sin\theta$, $dt = R\cos\theta\,d\theta$. When $t=-R$, we have $\theta=-\frac{\pi}{2}$; when $t=R$, we have $\theta=\frac{\pi}{2}$. So we get
$$\int_{-R}^R\sqrt{R^2-t^2}\,dt = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}R\cos^2\theta\,d\theta.$$
Now you can use the formula
$$\int \cos^2\theta\,d\theta = \frac{1}{2}\theta + \frac{1}{2}\sin\theta\cos\theta+C$$
(which can be found by using integration by parts, then replacing $\sin^2\theta$ with $1-\cos^2\theta$, and "solving" for the integral of the cosine squared) to get the desired result. 
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