Question

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

• A. $\underset{0}{\overset{\pi /4}{\int }}\tan xdx+\underset{\pi /4}{\overset{\pi /2}{\int }}\cot xdx$
• B. ${\int }_0^{\pi /4}{\tan }^{-1}xdx-{\int }_{\pi /4}^0{\cot }^{-1}xdx$
• C. $\frac{-2}{\pi }\underset{0}{\overset{\pi /2}{\int }}\ln \left(\sin x\right)dx$
• D. ${\int }_0^1{\tan }^{-1}xdx$

Answer 1

Answer: (A), (C)

The correct options are
A $\underset{0}{\overset{\pi /4}{\int }}\tan xdx+\underset{\pi /4}{\overset{\pi /2}{\int }}\cot xdx$
C $\frac{-2}{\pi }\underset{0}{\overset{\pi /2}{\int }}\ln \left(\sin x\right)dx$


$\text{Required area}=\underset{0}{\overset{\pi /4}{\int }}\tan ydy+\underset{\pi /4}{\overset{\pi /2}{\int }}\cot ydy$
$=\ln \sqrt{2}+\ln \sqrt{2}$
$=\ln 2$

${\int }_0^{\pi /4}{\tan }^{-1}xdx-{\int }_{\pi /4}^0{\cot }^{-1}xdx=\frac{{\pi }^2}{8}$

$\frac{-2}{\pi }\underset{0}{\overset{\pi /2}{\int }}\ln \sin xdx=I2I=\frac{-2}{\pi }{\int }_0^{\pi /2}\ln \sin 2xdx-\ln 2dx,2x=t\Rightarrow I=\ln 2$

$I=\int {\tan }^{-1}xdx=x{\tan }^{-1}x-\int \frac{x}{1+x^2}dx=x{\tan }^{-1}x-\frac{1}{2}\ln (1+x^2)\Rightarrow {\int }_0^1{\tan }^{-1}xdx=\frac{\pi }{4}-\frac{\ln 2}{2}$