Question

Evaluate : ${\displaystyle {\int }_0^{1}x{\tan }^{-1}x\cdot dx}$

Answer 1

Let I = ${\displaystyle {\int }_0^{1}x{\tan }^{-1}x\cdot dx}$

= ${\displaystyle {\int }_0^1\left({\tan }^{-1}x\right)\left(x\right)\cdot dx}$

= ${\displaystyle {[\left({\tan }^{-1}x\right)\int x\cdot dx]}_0^1-{\int }_0^1[\frac{d}{dx}\left({\tan }^{-1}x\right)\cdot \int x\cdot dx]\cdot dx}$

= ${\displaystyle {\left[\frac{x^2{\tan }^{-1}x}{2}\right]}_0^1-{\int }_0^1\frac{1}{1+x^2}\cdot \frac{x^2}{2}\cdot dx}$

= ${\displaystyle (\frac{1^2{\tan }^{-1}1}{2}-0)-\frac{1}{2}{\int }_0^1\frac{1+x^2-1}{1+x^2}\cdot dx}$

= ${\displaystyle \frac{\frac{\pi }{4}}{2}-\frac{1}{2}{\int }_0^1(1-\frac{1}{1+x^2})\cdot dx}$

= ${\displaystyle \frac{\pi }{8}-\frac{1}{2}{[x-{\tan }^{-1}\left(x\right)]}_0^1}$

= ${\displaystyle \frac{\pi }{8}-\frac{1}{2}[(1-{\tan }^{-1}1)-0]}$

= ${\displaystyle \frac{\pi }{8}-\frac{1}{2}(1-\frac{\pi }{4})}$

= ${\displaystyle \frac{\pi }{8}-\frac{1}{2}+\frac{\pi }{8}}$

= ${\displaystyle \frac{\pi }{4}-\frac{1}{2}}$.