Limit + Integration --> Limtegration!! πŸ˜†

Calculus Level 3

lim ⁑ t β†’ ∞ ∫ 0 t ( tan ⁑ βˆ’ 1 x ) 2 dx t 2 + 1 = ? \large \displaystyle\lim_{t\rightarrow \infty} \frac{\displaystyle\int_{0}^{t} (\tan^{-1}x)^2 \text{dx} }{\sqrt{t^2+1}} = ?

Ο€ 2 4 \frac{\pi^2}{4} Ο€ 2 2 \frac{\pi^2}{2} Ο€ 4 \frac{\pi}{4} Ο€ 2 \frac{\pi}{2}

Akhil Bansal by Akhil Bansal , 22, India

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1 solution

Joel Yip
Sep 8, 2015

Using L'Hopital's Rule

lim ⁑ t β†’ ∞ ∫ 0 t tan ⁑ βˆ’ 1 x 2 d x 1 + t 2 = lim ⁑ t β†’ ∞ ( tan ⁑ βˆ’ 1 t ) 2 t 1 + t 2 \lim _{ t\rightarrow \infty }{ \frac { \int _{ 0 }^{ t }{ { \tan ^{ -1 }{ x } }^{ 2 } } dx }{ \sqrt { 1+{ t }^{ 2 } } } } =\lim _{ t\rightarrow \infty }{ \frac { { \left( \tan ^{ -1 }{ t } \right) }^{ 2 } }{ \frac { t }{ \sqrt { 1+{ t }^{ 2 } } } } }

Then,

lim ⁑ t β†’ ∞ ( tan ⁑ βˆ’ 1 t ) 2 t 1 + t 2 = Ο€ 2 4 \lim _{ t\rightarrow \infty }{ \frac { { \left( \tan ^{ -1 }{ t } \right) }^{ 2 } }{ \frac { t }{ \sqrt { 1+{ t }^{ 2 } } } } } =\frac { { \pi }^{ 2 } }{ 4 }

5 Helpful 0 Interesting 0 Brilliant 1 Confused

Nicely done :) dt is missing do edit :)

RAJ RAJPUT - 5Β years, 9Β months ago

But how did u consider that numerator is infinity

Vaibhav Vinayaka - 2Β years, 8Β months ago

Shouldn't u use Leibniz's method for the numerator

Himanshu Borkar - 1Β year, 2Β months ago

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