Question

Evaluate
$\int \sqrt{1-x^{2}dx}$

Answer 1

$\int \sqrt{1-x^2}dx$

Let $x=\sin \theta \Rightarrow dx=\cos \theta d\theta$

$=\int \sqrt{1-{\sin }^2\theta }\cos \theta d\theta$

$=\int \sqrt{{\cos }^2\theta }\cos \theta d\theta$

$=\int {\cos }^2\theta d\theta$

$=\int {\cos }^2\theta d\theta$ since $\cos 2\theta =2{\cos }^2\theta -1\Rightarrow {\cos }^2\theta =\frac{1}{2}(1+\cos 2\theta )$

$=\frac{1}{2}\int (1+\cos 2\theta )d\theta$

$=\frac{1}{2}[\theta +\frac{\sin 2\theta }{2}]+c$

$=\frac{{\sin }^{-1}x}{2}+\frac{\sin 2\left({\sin }^{-1}x\right)}{4}+c$ where $\theta ={\sin }^{-1}x$