Question

The value of the integral $\underset{0}{\overset{1}{\int }}x{\cot }^{-1}\left(1-x^2+x^4\right)dx$ is :

• A. $\frac{\pi }{4}-\frac{1}{2}{\log }_{e}2$
• B. $\frac{\pi }{2}-{\log }_{e}2$
• C. $\frac{\pi }{2}-\frac{1}{2}{\log }_{e}2$
• D. $\frac{\pi }{4}-{\log }_{e}2$

Answer 1

Answer: (A)

$I={\int }_0^{1}x\tan \left(\frac{1}{1+x^2\left(x^2-1\right)}\right)dx$
$I={\int }_0^{1}x\left({\tan }^{-1}{x}^2-{\tan }^{-1}\left(x^2-1\right)\right)dx$
$x^2=t\Rightarrow 2xdx=dt$
$I=\frac{1}{2}{\int }_0^1\left({\tan }^{-1}t-{\tan }^{-1}\left(t-1\right)\right)dx$
$=\frac{1}{2}{\int }_0^1{\tan }^{-1}tdt-\frac{1}{2}{\int }_0^1{\tan }^{-1}\left(t-1\right)dt$
$=\frac{1}{2}{\int }_0^1{\tan }^{-1}tdt-\frac{1}{2}{\int }_0^1{\tan }^{-1}dt={\int }_0^1{\tan }^{-1}dt$
${\tan }^{-1}t=\theta \Rightarrow t=\tan \theta$
$dt={\sec }^2\theta d\theta$
${\int }_0^{\pi /4}\theta .{\sec }^2\theta d\theta$
$I={\left(\theta .\tan \theta \right)}_0^{\pi /4}-{\int }_0^{\pi /4}\tan \theta d\theta$
$=\left(\frac{\pi }{4}-0\right)-ln{\left(\sec \theta \right)}_0^{\pi /4}$
$=\frac{\pi }{4}-\left(\ell n\sqrt{2}-0\right)$
$=\frac{\pi }{4}-\frac{1}{2}\ell n2$