$\int\left(\frac{\tan \left(\frac{1}{x}\right)}{x}\right)^2 \mathrm{~d} x=$

$\int\left(\frac{\tan \left(\frac{1}{x}\right)}{x}\right)^2 \mathrm{~d} x=$
  1. $x-\tan x+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration
  2. $\frac{1}{x}-\tan \left(\frac{1}{x}\right)+c$, where $\mathrm{c}$ is a constant of integration.
  3. $\frac{1}{x}+\tan \left(\frac{1}{x}\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  4. $x+\tan x+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.

Solution

Let $\mathrm{I}=\int\left(\frac{\tan \left(\frac{1}{x}\right)}{x}\right)^2 \mathrm{~d} x$ Let $\frac{1}{x}=\mathrm{t} \Rightarrow \frac{1}{x^2} \mathrm{~d} x=-\mathrm{dt}$ $\begin{aligned} \therefore \quad \mathrm{I} & =-\int \tan ^2 \mathrm{t} d \mathrm{t} \\ & =\int\left(1-\sec ^2 \mathrm{t}\right) \mathrm{dt} \\ & =\mathrm{t}-\tan \mathrm{t}+\mathrm{c} \\ & =\frac{1}{x}-\tan \left(\frac{1}{x}\right)+\mathrm{c} \end{aligned}$

Asked in: MHT CET 2023 (12 May Shift 2)

Practice more Indefinite Integration questions on Aicharya