The art of Mathematics

$\begin{aligned} \int_1^2 \frac{dx}{\sqrt{2^2 - x^2}} & = \int_1^2 \frac{dx}{2\sqrt{1 - \frac{x^2}{4}}} \quad \quad \small \color{#3D99F6} {\text{Let } \frac{x}{2} = \sin{\theta} \space \Rightarrow x = 1, \theta = \frac{\pi}{6} \space \Rightarrow x = 2, \theta = \frac{\pi}{2} \space \Rightarrow \frac{dx}{2} = \cos{\theta} d\theta} \\ & = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\cos{\theta} d\theta}{\sqrt{1-\sin^2{\theta}}} \\ & = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\cos{\theta}}{\cos{\theta}}d \theta \\ & = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} d \theta = \left[ \theta \right]_{\frac{\pi}{6}}^{\frac{\pi}{2}} = \frac{\pi}{2} - \frac{\pi}{6} = \boxed{\dfrac{\pi}{3}} \end{aligned}$

We can find that $\displaystyle \int \dfrac{dx}{\sqrt{a^2-x^2}} = sin^{-1}\left(\frac{x}{a} \right) + C$ by making the simple substituition, $x = a\sin{\theta}$ Clearly $a = 2$ .Using the above formula we get, $\displaystyle \int^{2}_{1} \dfrac{dx}{\sqrt{2^2 - x^2}} = \left[sin^{-1}\left(\frac{x}{2} \right)\right]^{2}_{1} = \frac{\pi}{2} - \frac{\pi}{6} = \boxed{\dfrac{\pi}{3}}$

$\text{Proof}$ : Lets assume that $x=asin\theta$ then we have , $dx = acos\theta d\theta$ ,

Now we have, $\displaystyle \int \dfrac{dx}{\sqrt{a^2-x^2}} \Rightarrow \displaystyle \int \dfrac{acos\theta d\theta}{\sqrt{a^2 - a^2sin^2\theta}} \Rightarrow \displaystyle \int d\theta \Rightarrow \theta + C$ Since $x=asin\theta$ , we have $\theta = sin^{-1}\left(\dfrac{x}{a}\right)$ Therefore we can conclude that, $\displaystyle \int \dfrac{dx}{\sqrt{a^2-x^2}} = sin^{-1}\left(\frac{x}{a} \right) + C$