Question

Area bounded by curves $y={\cot }^{-1}x,$ $y={\tan }^{-1}x$ and $y-$ axis is equal to

• A. $\frac{-2}{\pi }{\int }_0^{\pi /2}\ln \left(\sin x\right)dx$
• B. ${\int }_0^{\pi /4}{\tan }^{-1}xdx+{\int }_{\pi /4}^{\pi /2}{\cot }^{-1}xdx$
• C. ${\int }_0^{\pi /4}\tan xdx+{\int }_{\pi /4}^{\pi /2}\cot xdx$
• D. ${\int }_0^{\pi /4}{\cot }^{-1}xdx+{\int }_{\pi /4}^{\pi /2}{\tan }^{-1}xdx$

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{-2}{\pi }{\int }_0^{\pi /2}\ln \left(\sin x\right)dx$
C ${\int }_0^{\pi /4}\tan xdx+{\int }_{\pi /4}^{\pi /2}\cot xdx$



Area $={\int }_0^{\pi /4}\left(\tan y\right)dy+{\int }_{\pi /4}^{\pi /2}\left(\cot y\right)dy$
$={\left[\ln |\sec y|\right]}_0^{\pi /4}+[\ln |\sin y|{]}_{\pi /4}^{\pi /2}$
$=[\ln \left[\sec \frac{\pi }{4}\right]-\ln |\sec 0|]+[\ln \left|\sin \frac{\pi }{2}\right|-\ln \left|\sin \frac{\pi }{4}\right|]$
$=[\ln \sqrt{2}-\ln 1]+\ln 1-\ln \frac{1}{\sqrt{2}}$
$=\ln \sqrt{2}+\ln \sqrt{2}=2\ln \sqrt{2}=\ln 2$

$\frac{-2}{\pi }{\int }_0^{\pi /2}\ln \left(\sin x\right)dx=\frac{-2}{\pi }[-\frac{\pi }{2}\ln 2]=\ln 2$