For a≥2, if the value of the definite integral ∫0∞dxa2+x−1x2 equals π10 ,then the value of a is

I=∫0∞dxx2+1x2+(a2−2)=∫0∞x2dxx4+(a2−2)x2+1, Put a2−2=k =∫0∞x2dxx4+kx2+1=12∫0∞(x2+1)+(x2−1)x4+kx2+1dx =12∫0∞1+(1x2)x2+(1x2)+k⏟I1dx+12∫0∞1−(1x2)x2+(1x2)+k⏟I2dxNow proceed, I1=π2a and I2=0,∴ I=π2a=π10⇒ a=5

Book Online Demo

Check Your IQ

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

For $a \geq 2,$ if the value of the definite integral $\int_{0}^{\infty} \frac{d x}{a^{2} + {(x - (\frac{1}{x}))}^{2}}$ equals $\frac{π}{10}$ ,then the value of a is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

answer is 5.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$I = \int_{0}^{\infty} \frac{d x}{x^{2} + \frac{1}{x^{2}} + (a^{2} - 2)} = \int_{0}^{\infty} \frac{x^{2} d x}{x^{4} + (a^{2} - 2) x^{2} + 1}, P u t a^{2} - 2 = k = \int_{0}^{\infty} \frac{x^{2} d x}{x^{4} + k x^{2} + 1} = \frac{1}{2} \int_{0}^{\infty} \frac{(x^{2} + 1) + (x^{2} - 1)}{x^{4} + k x^{2} + 1} d x = \frac{1}{2} \underset{I_{1}}{\underset{⏟}{\int_{0}^{\infty} \frac{1 + (\frac{1}{x^{2}})}{x^{2} + (\frac{1}{x^{2}}) + k}}} d x + \frac{1}{2} \underset{I_{2}}{\underset{⏟}{\int_{0}^{\infty} \frac{1 - (\frac{1}{x^{2}})}{x^{2} + (\frac{1}{x^{2}}) + k}}} d x$