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 $x=\frac{1}{t}$ \( \implies I = -\int_{2014}^\frac{1}{2014} \frac{tan^{-1}\frac{1}{t}}{t} dt =\int_\frac{1}{2014}^{2014} \frac{tan^{-1}\frac{1}{t}}{t} dt = \int_\frac{1}{2014}^{2014} \frac{tan^{-1}\frac{1}{x}}{x} dx \) $$$$
\(\mathbb{Therefore,}\) $$$$
\[2I=\int_\frac{1}{2014}^{2014} \frac{tan^{-1}x}{x} dx+\int_\frac{1}{2014}^{2014} \frac{tan^{-1}\frac{1}{x}}{x} dx\] $$$$
\[=\int_\frac{1}{2014}^{2014} \frac{tan^{-1}x+tan^{-1}\frac{1}{x}}{x} dx \] $$$$
\[=\int_\frac{1}{2014}^{2014} \frac{\frac{\pi}{2}}{x} dx \] $$$$
\[=\frac{\pi}{2}\int_\frac{1}{2014}^{2014} \frac{dx}{x} \] $$$$
\[=\frac{\pi}{2}[\log x]_{\frac{1}{2014}}^{2104} \] $$$$
\[=\frac{\pi}{2}\big[\log 2014-\log \frac{1}{2014}\big]\] $$$$
\[=\frac{\pi}{2}\log 2014^2\] $$$$
\[=\pi\log 2014\] $$$$
No comments:
Post a Comment