The value of the integral 1 / 2 2 tan - 1 x x d x is equal to [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
$The value of the integral \int_{1 / 2}^{2} \frac{\tan^{- 1} x}{x} d x is equal to$ [2023]

1 $\frac{1}{2} \log_{e} 2$

2 $π \log_{e} 2$

3 $\frac{π}{4} \log_{e} 2$

4 $\frac{π}{2} \log_{e} 2$

Ans.
(4)

We have the integral as follows

$I = \int_{1 / 2}^{2} \frac{\tan^{- 1} x}{x} d x \dots (1)$

By substituting $x = \frac{1}{t}$ and $d x = - \frac{1}{t^{2}} d t$ in equation (1),

$I = - \int_{2}^{1 / 2} \frac{\tan^{- 1} \frac{1}{t}}{\frac{1}{t}} \cdot \frac{1}{t^{2}} \cdot d t$

$I = - \int_{2}^{1 / 2} \frac{\tan^{- 1} \frac{1}{t}}{t} d t = \int_{1 / 2}^{2} \frac{\cot^{- 1} t}{t} d t = \int_{1 / 2}^{2} \frac{\cot^{- 1} x}{x} d x \dots (2)$

On adding equation (1) and (2), we have

$2 I = \int_{1 / 2}^{2} \frac{\tan^{- 1} x + \cot^{- 1} x}{x} d x;$ $2 I = \frac{π}{2} \int_{1 / 2}^{2} \frac{d x}{x} = \frac{π}{2} {[\log x]}_{1 / 2}^{2}$

$2 I = \frac{π}{2} [\log_{e} 2 - \log_{e} \frac{1}{2}] = π \log_{e} 2$

$\Rightarrow$ $2 I = {πlog}_{e} 2$ $∴$ $I = \frac{π}{2} \log_{e} 2$

Hence, option (4) is the correct answer.

Want Us to Email you About Special Offers & Updates?