Question

If $2{\int }_0^1{\tan }^{-1}xdx={\int }_0^1{\cot }^{-1}(1-x+x^2)dx,$ then ${\int }_0^1{\tan }^{-1}(1-x+x^2)dx$ is equal to :

• A. $\log 4$
• B. $\frac{\pi }{2}\log 2$
• C. $\log 2$
• D. $\frac{\pi }{2}-\log 4$

Answer 1

Answer: (C)

$2\underset{0}{\overset{1}{\int }}ta{n}^{-1}xdx=$$\underset{0}{\overset{1}{\int }}$ $\left(\frac{\pi }{2}-ta{n}^{-1}\left(1-x+x^2\right)\right)dx$
$2\underset{0}{\overset{1}{\int }}ta{n}^{-1}xdx=$$\underset{0}{\overset{1}{\int }}$$\frac{\pi }{2}dx-$$2\underset{0}{\overset{1}{\int }}ta{n}^{-1}(1-x+x^2)dx$
$2\underset{0}{\overset{1}{\int }}ta{n}^{-1}(1-x+x^2)dx=$$\frac{\pi }{2}-2$$\underset{0}{\overset{1}{\int }}ta{n}^{-1}xdx$
Let
$I_1=\underset{0}{\overset{1}{\int }}ta{n}^{-1}xdx$
$={\left[\left(ta{n}^{-1}x\right)x\right]}_0^{{}^1}-$ $\underset{0}{\overset{1}{\int }}$$\frac{1}{1+x^2}xdx$
$=\frac{\pi }{4}-$$\underset{0}{\overset{1}{\int }}$$\frac{x}{1+x^2}dx$
$=\frac{\pi }{4}-\frac{1}{2}In2$
By equation____(1)
$\frac{\pi }{2}-2\left[\frac{\pi }{4}-\frac{1}{2}In\right]=In2$