Evaluate ∫0∞ tan−1⁡xx1+x2dx

I=∫0∞ tan−1⁡xx1+x2dx Put x=tan⁡t so that dx=sec2⁡tdt. Then, I=∫0π/2 ttan⁡t⋅sec2⁡t⋅sec2⁡tdt=∫0π/2 tcot⁡tdt=[tlog⁡sin⁡t]0π/2−∫0π/2 1⋅log⁡sin⁡tdt=−∫0π/2 log⁡sin⁡tdt=−−π2log⁡2=π2log⁡2

Evaluate ∫0∞ tan−1⁡xx1+x2dx

I=∫0∞ tan−1⁡xx1+x2dx Put x=tan⁡t so that dx=sec2⁡tdt. Then, I=∫0π/2 ttan⁡t⋅sec2⁡t⋅sec2⁡tdt=∫0π/2 tcot⁡tdt=[tlog⁡sin⁡t]0π/2−∫0π/2 1⋅log⁡sin⁡tdt=−∫0π/2 log⁡sin⁡tdt=−−π2log⁡2=π2log⁡2

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$Evaluate \int_{0}^{\infty} \frac{\tan^{- 1} x}{x (1 + x^{2})} d x$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{2}{π} \log 2$

$\frac{π}{2} \log 2$

$\frac{1}{2} \log 2$

$\frac{1}{π} \log 2$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} I = \int_{0}^{\infty} \frac{\tan^{- 1} x}{x (1 + x^{2})} d x Put x = \tan t so that d x = \sec^{2} t d t . Then, \\ I = \int_{0}^{π / 2} \frac{t}{\tan t \cdot \sec^{2} t} \cdot \sec^{2} t d t = \int_{0}^{π / 2} t \cot t d t = [t \log \sin t]_{0}^{π / 2} - \int_{0}^{π / 2} 1 \cdot \log \sin t d t \\ = - \int_{0}^{π / 2} \log \sin t d t = - (- \frac{π}{2} \log 2) = \frac{π}{2} \log 2 \end{array}$