Question

Area bounded by the curve $y=si{n}^{-1}x,y-axis$ and $y=co{s}^{-1}x$ is equal to

• A. $(2+\sqrt{2})$
• B. $(2-\sqrt{2})$
• C. $(1+\sqrt{2})$
• D. $(\sqrt{2}-1)$

Answer 1

The correct option is D $(\sqrt{2}-1)$

Given,

$y={\sin }^{-1}x,y={\cos }^{-1}x,y-axis$

$\Rightarrow \sin y=x,\cos y=x$

$\Rightarrow \sin y=\cos y\Rightarrow \tan y=1$

$\therefore y={\tan }^{-1}1=\frac{\pi }{4}$

$Area={\int }_0^{\frac{\pi }{4}}\cos \left(y\right)-\sin \left(y\right)dy$

$={\int }_0^{\frac{\pi }{4}}\cos \left(y\right)dy-{\int }_0^{\frac{\pi }{4}}\sin \left(y\right)dy$

$=\frac{1}{\sqrt{2}}-(-\frac{1}{\sqrt{2}}+1)$

$=\sqrt{2}-1$