Question

$\int \sqrt{4-x^2}dx$ is equals to

Answer 1

$I=\int \sqrt{4-x^2}dx$

Take $x=2\sin \theta \Rightarrow dx=2\cos \theta d\theta$

$\sqrt{4-x^2}=\sqrt{4-4{\sin }^2\theta }=\sqrt{4(1-{\sin }^2\theta )}=2\sqrt{{\cos }^2\theta }=2\cos \theta$

$\int \sqrt{4-x^2}dx$

$=\int 2\cos \theta .2\cos \theta d\theta$

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

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

$=2(\theta +\frac{\sin 2\theta }{2})+c$

$=2({\sin }^{-1}x/2+\frac{\sin 2{\sin }^{-1}x/2}{2})+c$ since $x=2\sin \theta \Rightarrow \theta ={\sin }^{-1}x/2$