Question

Find the area (sq.units) bounded by $y={\sin }^{-1}x$ and $y={\cos }^{-1}x$ and x-axis.
(Find answer correct upto two decimal places)

Answer 1

We have,

$y={\sin }^{-1}x$

$y={\cos }^{-1}x$

$y=0$

Area enclosed by these curves,

$={\int }_0^{1/\sqrt{2}}{\sin }^{-1}xdx+{\int }_{1/\sqrt{2}}^1{\cos }^{-1}xdx$

$={[x{\sin }^{-1}x+\sqrt{1-x^2}]}_0^{1/\sqrt{2}}+{[x{\cos }^{-1}x-\sqrt{1-x^2}]}_{1/\sqrt{2}}^1$

$=[(\frac{1}{\sqrt{2}}{\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)+\sqrt{1-\frac{1}{2}})-(0+\sqrt{1-0})]+[({\cos }^{-1}\left(1\right)-\sqrt{1-1})-(\frac{1}{\sqrt{2}}{\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)-\sqrt{1-\frac{1}{2}})]$

$=[(\frac{1}{\sqrt{2}}{\sin }^{-1}\left(\sin \frac{\pi }{4}\right)+\sqrt{\frac{1}{2}})-1]+[\left({\cos }^{-1}\left(\cos 0\right)\right)-(\frac{1}{\sqrt{2}}{\cos }^{-1}\left(\cos \frac{\pi }{4}\right)-\sqrt{\frac{1}{2}})]$

$=\frac{\pi }{4\sqrt{2}}+\frac{1}{\sqrt{2}}-1-\frac{\pi }{4\sqrt{2}}+\frac{1}{\sqrt{2}}$

$=\sqrt{2}-1$

$=0.41$sq.units

Hence, this is the answer.