Question

Find the area bounded by $y={\cos }^{-1}x,y={\sin }^{-1}x$ and $y-$axis

• A. $(2-\sqrt{2})$ sq. units
• B. $(\sqrt{2}-2)$ sq. units
• C. $2\sqrt{2}$ sq. units
• D. $\sqrt{2}$ sq. units

Answer 1

The correct option is A $(2-\sqrt{2})$ sq. units

$y={\cos }^{-1}x$ and $y={\sin }^{-1}x$ intersect at $P\equiv (\frac{1}{\sqrt{2}},\frac{\pi }{4})$
Hence, required area is,
${\int }_0^{\frac{1}{\sqrt{2}}}({\cos }^{-1}x-{\sin }^{-1}x)dx={\int }_0^{\frac{1}{\sqrt{2}}}{\cos }^{-1}xdx-{\int }_0^{\frac{1}{\sqrt{2}}}{\sin }^{-1}xdx$
$={\left[x{\cos }^{-1}x\right]}_0^{\frac{1}{\sqrt{2}}}-{\int }_0^{\frac{1}{\sqrt{2}}}x.\frac{-1}{\sqrt{1-x^2}}dx-{\left[x{\sin }^{-1}x\right]}_0^{\frac{1}{\sqrt{2}}}+{\int }_0^{\frac{1}{\sqrt{2}}}x.\frac{1}{\sqrt{1-x^2}}dx$

$=\left[\frac{1}{\sqrt{2}}{\cos }^{-1}\frac{1}{\sqrt{2}}\right]-\left[\frac{1}{\sqrt{2}}{\sin }^{-1}\frac{1}{\sqrt{2}}\right]+{\int }_0^{\frac{1}{\sqrt{2}}}2x.\frac{1}{\sqrt{1-x^2}}dx$

$={\int }_0^{\frac{1}{\sqrt{2}}}2x.\frac{1}{\sqrt{1-x^2}}dx$

Assuming $1-x^2=t^2$

Differentiating both sides, we get

$-2xdx=2tdt$

Substituting in the integral, we get

$={\int }_{\frac{1}{\sqrt{2}}}^{1}2dt$

$=2-\sqrt{2}$