Question

If f(x) = $\int\limits^x_1 \frac{tan^{-1}(t)}{t} \, dt$ x ∈ R, then find the value of f(e)² - f(1/e²)​

Answer 1

Step-by-step explanation:

Given:-

• $\sf f(x) = \int \limits ^{x} _{1}\dfrac{ {tan}^{ - 1} (t)}{t}dt$

To Find:-

• $\sf The \: value \: of : \: f( {e}^{2} ) - f \bigg( \dfrac{1}{ {e}^{2} } \bigg)$

Solution:-

$\sf f(x) = \int \limits ^{x} _{1}\dfrac{ {tan}^{ - 1} (t)}{t}dt$

$\sf \therefore f \bigg( \dfrac{1}{x} \bigg ) = \int \limits ^{1/x} _{1}\dfrac{ {tan}^{ - 1} (t)}{t}dt$

$\sf Put \: \: t = \dfrac{1}{u}$

$\sf \therefore dt = - \dfrac{du}{ {u}^{2} }$

$\sf f \bigg( \dfrac{1}{x} \bigg ) = \int \limits ^{x} _{1}\dfrac{ {tan}^{ - 1} \bigg(\dfrac{1}{u} \bigg) }{\dfrac{1}{u}} \bigg( - \dfrac{1}{ {u}^{2} } \bigg) du$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = - \int \limits ^{x} _{1}\dfrac{ {tan}^{ - 1} \bigg(\dfrac{1}{u} \bigg) }{u} du$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = - \int \limits ^{x} _{1}\dfrac{ {cot}^{ - 1}(u)}{u} du$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = - \int \limits ^{x} _{1}\dfrac{ {cot}^{ - 1}(t)}{t} dt$

$\sf Now, \: f(x) - f\bigg(\dfrac{1}{x}\bigg)$

$\sf = \int \limits ^{x} _{1}\dfrac{ {tan}^{ - 1} (t)}{t}dt - \int \limits ^{x} _{1}\dfrac{ {cot}^{ - 1}(t)}{t} dt$

$\sf = \int \limits ^{x} _{1}\dfrac{ {tan}^{ - 1} (t)}{t}dt - \bigg( - \int \limits ^{x} _{1}\dfrac{ {cot}^{ - 1}(t)}{t} dt \bigg)$

$\sf = \int \limits ^{x} _{1}\dfrac{ {tan}^{ - 1} (t) + {cot}^{ - 1} (t)}{t}dt$

$\sf = \int \limits ^{x} _{1} \dfrac{\pi}{2} \times \dfrac{1}{t} dt$

$\sf = \dfrac{\pi}{2} log(x)$

$\underline{ \boxed{ \therefore\sf \: f( {e}^{2} ) - f \bigg( \dfrac{1}{ {e}^{2} } \bigg) = \frac{\pi}{2} log_{e}( {e}^{2}) = \pi}}$