Q.Evaluate: $\displaystyle\int_0^2 \dfrac{1}{\sqrt{4 - x^2}}\, dx$

With $a = 2$, $\int_0^2 \frac{dx}{\sqrt{4-x^2}} = \left[\sin^{-1}\frac{x}{2}\right]_0^2 = \sin^{-1}1 - \sin^{-1}0 = \frac{\pi}{2}$.

Use the standard result $\displaystyle\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\!\left(\frac{x}{a}\right) + c$ with $a^2 = 4$, i.e. $a = 2$:

$\int_0^2 \frac{dx}{\sqrt{4 - x^2}} = \left[\sin^{-1}\!\left(\frac{x}{2}\right)\right]_0^2.$

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