Question

The value of the integral ${\int }_0^{1}x{\cot }^{-1}(1-x^2+x^4)dx$ i

• 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)

$\frac{\pi }{4}-\frac{1}{2}{\log }_{e}2$

$I={\int }_0^{1}x\tan \left(\frac{1}{1+x^2(x^2-1)}\right)dx$

$x^2=t\Rightarrow 2xdx=dt$

$I=\frac{1}{2}{\int }_0^1({\tan }^{-1}t-{\tan }^{-1}(t-1))dx=\frac{1}{2}{\int }_0^1\left({\tan }^{-1}t\right)dt-\frac{1}{2}{\int }_0^1\left({\tan }^{-1}(t-1)\right)dt$

$={\int }_0^1\left({\tan }^{-1}t\right)dt$

${\tan }^{-1}t=\theta \Rightarrow t=\tan \theta dt={\sec }^2\theta d\theta$

${\int }_0^{\pi /4}\theta .{\sec }^2\theta d\theta$

$I={\left|\left(\theta \tan \theta \right)\right|}_0^{\pi /4}-{\int }_0^1\left(\tan \theta \right)d\theta =(\frac{\pi }{4}-0)-\ln {\left|\sec \theta \right|}_0^{\pi /4}=\frac{\pi }{4}-(\ln \sqrt{2}-0)=\frac{\pi }{4}-\frac{1}{2}\ln 2$