Question

The value of the integral $\int\limits^{1}_{0} 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)

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