## Wednesday, May 27, 2015

### Indian Statistical Institute B.Math & B.Stat : Integration

Indian Statistical Institute B.Math & B.Stat Evaluate: $\int_\frac{1}{2014}^{2014} \frac{tan^{-1}x}{x} dx$  Let $I=\int_\frac{1}{2014}^{2014} \frac{tan^{-1}x}{x} dx$, Put $x=\frac{1}{t}$ $\implies I = -\int_{2014}^\frac{1}{2014} \frac{tan^{-1}\frac{1}{t}}{t} dt =\int_\frac{1}{2014}^{2014} \frac{tan^{-1}\frac{1}{t}}{t} dt = \int_\frac{1}{2014}^{2014} \frac{tan^{-1}\frac{1}{x}}{x} dx$  $\mathbb{Therefore,}$  $2I=\int_\frac{1}{2014}^{2014} \frac{tan^{-1}x}{x} dx+\int_\frac{1}{2014}^{2014} \frac{tan^{-1}\frac{1}{x}}{x} dx$  $=\int_\frac{1}{2014}^{2014} \frac{tan^{-1}x+tan^{-1}\frac{1}{x}}{x} dx$  $=\int_\frac{1}{2014}^{2014} \frac{\frac{\pi}{2}}{x} dx$  $=\frac{\pi}{2}\int_\frac{1}{2014}^{2014} \frac{dx}{x}$  $=\frac{\pi}{2}[\log x]_{\frac{1}{2014}}^{2104}$  $=\frac{\pi}{2}\big[\log 2014-\log \frac{1}{2014}\big]$  $=\frac{\pi}{2}\log 2014^2$  $=\pi\log 2014$