An Arctangent Integral

Calculus Level 5

The following integral evaluates to:

∫ 0 1 tan ⁡ − 1 x 2 d x = 1 2 ln ⁡ ( 1 + a 1 − a ) a + π 2 ( 1 2 − a ) \int_{0}^{1} \tan^{-1} {x^2} dx = \frac{1}{2} \ln{ \bigg( \frac{1+a}{1-a}} \bigg)^{a} + \frac{\pi}{2} \bigg( \frac{1}{2}-a \bigg)

Where a > 1 2 a > \frac{1}{2} . What is a a ?


The answer is 0.707106781.

Denis Kartachov by Denis Kartachov , 24, Canada

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2 solutions

Naren Bhandari
Aug 16, 2018

Call the integral I = ∫ arc tan ⁡ x 2 d x = x arc tan ⁡ x 2 − ∫ 2 x 2 1 + x 4 d x ⏟ Integration by parts ∴ I = x tan ⁡ − 1 x 2 − { 1 2 tan ⁡ − 1 ( x 2 − 1 2 x ) + 1 2 2 ln ⁡ [ x 2 + 1 − 2 x x 2 + 1 + 2 x ] } ⏟ See notes + C \begin{aligned} I & = \int \text{arc}\tan x^2 \,dx \\ & = \underbrace{x \text{arc} \tan x^2 - \int \dfrac{2x^2}{1+x^4}\,dx}_{\text{Integration by parts}} \\ \therefore I & = x\tan ^{-1} x^2 -\underbrace{\left\{\dfrac{1}{\sqrt 2} \tan^{-1}\left(\dfrac{x^2-1}{ 2\sqrt x}\right) +\dfrac{1}{2\sqrt 2 }\ln\left[\dfrac{x^2+1-\sqrt 2 x }{x^2+1+\sqrt 2 x }\right]\right\}}_{\text{See notes}}+ C\end{aligned} Setting the limits we can find ∫ 0 1 tan ⁡ − 1 x 2 d x = π 4 − { π 2 2 + π 2 2 + 1 2 2 ln ⁡ 2 − 2 2 + 2 } = − 1 2 2 ln ⁡ 2 − 2 2 + 2 − π 4 2 ( 2 − 2 ) = 1 2 ( ln ⁡ 1 + 1 2 1 − 1 2 ) 1 2 + π 2 ( 1 2 − 1 2 ) \begin{aligned} \int_{0}^{1} \tan ^{-1} x^2\,dx & = \dfrac{\pi}{4}- \left\{ \dfrac{\pi }{2\sqrt 2 }+ \dfrac{\pi}{2\sqrt 2}+\dfrac{1}{2\sqrt2}\ln \dfrac{2-\sqrt 2}{2 +\sqrt 2}\right\} \\ & =-\dfrac{1}{2\sqrt 2} \ln \dfrac{2-\sqrt 2}{2 +\sqrt 2}- \dfrac{\pi}{4\sqrt 2}\left(\sqrt 2 -2\right)\\ & = \dfrac{1}{2}\left( \ln \dfrac{1+\frac{1}{\sqrt 2}}{1-\frac{1}{\sqrt 2}}\right)^{\frac{1}{\sqrt2}}+ \dfrac{\pi}{2}\left(\frac{1}{2}-\frac{1}{\sqrt 2} \right) \end{aligned} Therefore, required of a = 1 2 ≈ 0.707 a =\frac{1}{\sqrt 2} \approx \boxed{0.707} .

Note:

∫ x 2 1 + x 4 d x = 1 2 ∫ ( x 2 + 1 1 + x 4 + x 2 − 1 1 + x 4 ) d x = 1 2 ∫ ( 1 + 1 x 2 x 2 + 1 x 2 + 1 − 1 x 2 x 2 + 1 x 2 ) d x = 1 2 ∫ ( 1 + 1 x 2 ( x − 1 x ) 2 + 2 + 1 − 1 x 2 ( x + 1 x ) 2 − 2 ) d x \begin{aligned} \int \dfrac{x^2}{1+x^4}\,dx & =\dfrac{1}{2}\int\left(\dfrac{x^2+1}{1+x^4}+ \dfrac{x^2-1}{1+x^4} \right)\,dx\\& = \dfrac{1}{2}\int\left(\dfrac{1+\frac{1}{x^2} }{x^2+\frac{1}{x^2} }+ \dfrac{1-\frac{1}{x^2} }{x^2+\frac{1}{x^2} } \right)\,dx \\ & =\dfrac{1}{2}\int\left(\dfrac{1+\frac{1}{x^2} }{\,(x-\frac{1}{x})^2 +2 }+ \dfrac{1-\frac{1}{x^2} }{{\,(x+\frac{1}{x})^2 -2} } \right)\,dx\end{aligned} Substitute x − 1 x = t ⟹ 1 + 1 x 2 = d t x -\frac{1}{x} = t \implies 1+\frac{1}{x^2}=\,dt and x + 1 x = u ⟹ 1 − 1 x 2 = d u x +\frac{1}{x} = u\implies 1-\frac{1}{x^2}=\,du . We have then ∫ 2 x 2 1 + x 4 = 2 2 ∫ ( d t t 2 + 2 + d u u 2 − 2 ) = ( 1 2 tan ⁡ − 1 ( x 2 − 1 2 x ) + 1 2 2 ln ⁡ [ x 2 + 1 − 2 x x 2 + 1 + 2 x ] ) + C \begin{aligned} \int \dfrac{2x^2}{1+x^4} & = \dfrac{2}{2}\int \left( \dfrac{\,dt}{ t^2 +2} +\dfrac{\,du}{u^2-2}\right) \\& = \left(\dfrac{1}{\sqrt 2}\tan^{-1} \left(\dfrac{x^2-1}{2\sqrt x}\right)+\dfrac{1}{2\sqrt 2}\ln\left[\dfrac{x^2+1-\sqrt 2 x}{x^2+1+\sqrt 2 x}\right]\right)+C \end{aligned}

7 Helpful 0 Interesting 2 Brilliant 0 Confused

Love the two substitutions instead of the partial fractions method! Well done

Denis Kartachov - 2 years, 9 months ago
Chew-Seong Cheong
Aug 17, 2018

I = ∫ 0 1 tan ⁡ − 1 x 2 d x By integration by parts = x tan ⁡ − 1 x 2 ∣ 0 1 − ∫ 0 1 2 x 2 x 4 + 1 d x = π 4 − ∫ 0 1 2 x 2 ( x 2 − 2 x + 1 ) ( x 2 + 2 x + 1 ) d x By partial fraction decomposition = π 4 − ∫ 0 1 1 2 ( x x 2 − 2 x + 1 − x x 2 + 2 x + 1 ) d x \begin{aligned} I & = \int_0^1 \tan^{-1} x^2 \ dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = x \tan^{-1} x^2 \bigg|_0^1 - \int_0^1 \frac {2x^2}{x^4+1} dx \\ & = \frac \pi 4 - \color{#3D99F6} \int_0^1 \frac {2x^2}{(x^2-\sqrt 2x+1)(x^2+\sqrt 2x +1)} dx & \small \color{#3D99F6} \text{By partial fraction decomposition} \\ & = \frac \pi 4 - \color{#3D99F6} \int_0^1 \frac 1{\sqrt 2}\left(\frac x{x^2-\sqrt 2x+1} - \frac x{x^2+\sqrt 2x+1}\right) dx \end{aligned}

= π 4 − 1 2 2 ∫ 0 1 ( 2 x − 2 x 2 − 2 x + 1 − 2 x + 2 x 2 + 2 x + 1 ) d x − 1 2 ∫ 0 1 ( 1 x 2 − 2 x + 1 + 1 x 2 + 2 x + 1 ) d x = π 4 − 1 2 2 ln ⁡ ( x 2 − 2 x + 1 x 2 + 2 x + 1 ) + 1 2 ( tan ⁡ − 1 ( 1 − 2 x ) − tan ⁡ − 1 ( 1 + 2 x ) ) ∣ 0 1 = π 4 + 1 2 2 ln ⁡ ( 2 + 2 2 − 2 ) + 1 2 tan ⁡ − 1 − 2 x 1 − x 2 ∣ 0 1 = π 4 + 1 2 2 ln ⁡ ( 1 + 1 2 1 − 1 2 ) − π 2 2 = 1 2 ln ⁡ ( 1 + 1 2 1 − 1 2 ) 1 2 + π 2 ( 1 2 − 1 2 ) \begin{aligned} \ \ & = \frac \pi 4 - \frac 1{2\sqrt 2} \int_0^1 \left(\frac {2x-\sqrt 2}{x^2-\sqrt 2x+1} - \frac {2x+\sqrt 2}{x^2+\sqrt 2x+1}\right) dx - \frac 12 \int_0^1 \left(\frac 1{x^2-\sqrt 2x+1} + \frac 1{x^2+\sqrt 2x+1}\right) dx \\ & = \frac \pi 4 - \frac 1{2\sqrt 2} \ln \left(\frac {x^2-\sqrt 2x+1}{x^2+\sqrt 2x+1}\right) + \frac 1{\sqrt 2}\left(\tan^{-1} (1-\sqrt 2x) - \tan^{-1} (1+\sqrt 2x) \right) \bigg|_0^1 \\ & = \frac \pi 4 + \frac 1{2\sqrt 2} \ln \left(\frac {2+\sqrt 2}{2-\sqrt 2}\right) + \frac 1{\sqrt 2} \tan^{-1} \frac {-\sqrt 2x}{1-x^2} \bigg|_0^1 \\ & = \frac \pi 4 + \frac 1{2\sqrt 2} \ln \left(\frac {1+\frac 1{\sqrt 2}}{1-\frac 1{\sqrt 2}}\right) - \frac \pi {2\sqrt 2} \\ & = \frac 12 \ln \left(\frac {1+\frac 1{\sqrt 2}}{1-\frac 1{\sqrt 2}}\right)^{\frac 1{\sqrt 2}} + \frac \pi 2\left(\frac 12 - \frac 1{\sqrt 2}\right) \end{aligned}

Therefore a = 1 2 ≈ 0.707 a = \dfrac 1{\sqrt 2} \approx \boxed{0.707} .

4 Helpful 0 Interesting 1 Brilliant 0 Confused

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