Question

Integrate $\int \sqrt{1-t^2}dt$.

Answer 1

$\int \sqrt{1-t^2}dt$

let $t=sin\theta \Rightarrow \theta =si{n}^{-1}\left(t\right)$ and $cos\theta =\sqrt{1-si{n}^2\theta }$

$=\sqrt{1-t^2}$

$dt=cos\theta d\theta$

$\int \sqrt{1-si{n}^2\theta }cos\theta d\theta =\int cos\theta .cos\theta .d\theta$

$=\int co{s}^2\theta d\theta$

$=\int \left(\frac{1+cos2\theta }{2}\right)d\theta$

$=\frac{1}{2}\int d\theta +\frac{1}{2}\int cos2\theta d\theta$

$=\frac{1}{2}\theta +\frac{1}{2}.\frac{sin2\theta }{2}+C$

$=\frac{\theta }{2}+\frac{sin2\theta }{4}+c$

$=\frac{1}{2}si{n}^{-1}\left(t\right)+\frac{2sincos\theta }{4}+c$

$=\frac{1}{2}si{n}^{-1}\left(t\right)+\frac{1}{2}.t\sqrt{1-t^2}+c$

$\int \sqrt{1-t^2}dt=\frac{1}{2}si{n}^{-1}\left(t\right)+\frac{t}{2}\sqrt{1-t^2}+c$ (where $t=\sin \theta$)