<s>[INST] Proof that $\int_{-1}^{1}\sqrt{1-x^2}dx=\frac{\pi}{2}$? In my textbook appears that $\displaystyle\int_{-1}^{1}\sqrt{1-x^2}dx=\frac{\pi}{2}$
But where does this equation come from?
 [/INST] Not the "cleverest" method, like the above, - but works!
$$\int_{-1}^1 \sqrt{1-x^2} \ dx$$
To compute that integral, one may substitute $x=\sin{t}$, and get:
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{1-\sin^2{t}} \ d(\sin{t})=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos{t}\cos{t} \ dt=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2{t} \ dt$$
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2{t} \ dt=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1+\cos{2t}}{2} \ dt=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{dt}{2}+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos{2t}}{2}=$$
$$= \frac{\pi}{2}+\frac{1}{2}\sin{t}\cos{t}{\huge{|}}_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=\frac{\pi}{2}+0=\frac{\pi}{2}$$
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