prove-that-Nice-Integral-0-1-tan-1-x-3-2-x-2-dx-pi-3-ln-7-4-3-4-m-n-

Question Number 165328 by mnjuly1970 last updated on 30/Jan/22

p r o v e t h a t

N i c e I n t e g r a l

\boldsymbol ϕ = \int_{0}^{1} \frac{{t a n}^{- 1} (x^{\frac{3}{2}})}{x^{2}} d x = \frac{π + \sqrt{3} l n (7 + 4 \sqrt{3})}{4} m . n

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Answered by Lordose last updated on 30/Jan/22

Φ = \int_{0}^{1} \frac{\tan^{- 1} (x^{\frac{3}{2}})}{x^{2}} dx \overset{\boldsymbol IBP}{=} - \frac{\tan^{- 1} (x^{\frac{3}{2}})}{x} ∣_{0}^{1} + \frac{3}{2} \int_{0}^{1} \frac{x^{\frac{1}{2} - 1}}{1 + x^{3}} dx

Φ = - \frac{\boldsymbol π}{4} + \frac{3}{2} \int_{0}^{1} \frac{1}{\sqrt{x} (1 + x^{3})} dx

Φ = - \frac{\boldsymbol π}{4} + 3 \int_{0}^{1} \frac{1}{(1 + x^{6})} dx

Φ = - \frac{\boldsymbol π}{4} + 3 \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n - 1} \int_{0}^{1} x^{6 n - 6} dx

Φ = - \frac{\boldsymbol π}{4} + 3 \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{6 n - 5}

Φ = - \frac{\boldsymbol π}{4} + 3 (\frac{\boldsymbol π}{6} + \frac{\coth^{- 1} (\sqrt{3})}{\sqrt{3}})

Φ = \frac{\boldsymbol π}{4} + \sqrt{3} \coth^{- 1} (\sqrt{3})

Answered by mnjuly1970 last updated on 30/Jan/22

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\boldsymbol ϕ \overset{i . b . p}{=} {[\frac{- 1}{x} {t a n}^{- 1} (x^{\frac{3}{2}})]}_{0}^{1} + \frac{3}{2} \int_{0}^{1} \frac{1}{\sqrt{x} . (1 + x^{3})} d x

= - \frac{π}{4} + \frac{3}{2} Ω w h e r e Ω = \int_{0}^{1} \frac{d x}{\sqrt{x} (1 + x^{3})}

Ω \overset{\sqrt{x} = t}{=} \int_{0}^{1} \frac{2}{1 + x^{6}} d x = \int_{0}^{1} \frac{x^{4} + 1 - (x^{4} - 1)}{1 + x^{6}} d x

= \int_{0}^{1} \frac{(x^{4} - x^{2} + 1) + x^{2}}{1 + x^{6}} d x + \int_{0}^{1} \frac{1 - x^{2}}{1 - x^{2} + x^{4}} d x

= \frac{π}{4} + \int_{0}^{1} \frac{3 x^{2}}{1 + {(x^{3})}^{2}} d x - \int_{0}^{1} \frac{1 - x^{- 2}}{{(x + x^{- 1})}^{2} - 3} d x

= \frac{π}{4} + \frac{π}{4} + \int_{2}^{\infty} \frac{d x}{x^{2} - 3}

= \frac{π}{2} + \int_{2}^{\infty} \frac{d x}{(x - \sqrt{3}) (x + \sqrt{3})}

= \frac{π}{2} + \frac{1}{2 \sqrt{3}} {{[l n (\frac{x - \sqrt{3}}{x + \sqrt{3}})]}_{2}^{\infty}}

= \frac{π}{2} + \frac{1}{2 \sqrt{3}} l n (\frac{2 + \sqrt{3}}{2 - \sqrt{3}})

= \frac{π}{2} + \frac{1}{2 \sqrt{3}} l n (7 + 4 \sqrt{3})

∴ \boldsymbol ϕ = \frac{π}{4} + \frac{\sqrt{3}}{4} l n (7 + 4 \sqrt{3}) m . n

Answered by Ar Brandon last updated on 30/Jan/22

\boldsymbol ϕ = \int_{0}^{1} \frac{\tan^{- 1} (x^{\frac{3}{2}})}{x^{2}} d x = - {[\frac{1}{x} \tan^{- 1} (x^{\frac{3}{2}})]}_{0}^{1} + \frac{3}{2} \int_{0}^{1} \frac{x^{- \frac{1}{2}}}{1 + x^{3}} d x

= - \frac{π}{4} + 3 \int_{0}^{1} \frac{d t}{1 + t^{6}} = - \frac{π}{4} + 3 \int_{0}^{1} \frac{d t}{(t^{2} + 1) (t^{4} - t^{2} + 1)}

= - \frac{π}{4} + \int_{0}^{1} (\frac{1}{t^{2} + 1} - \frac{t^{2} - 2}{t^{4} - t^{2} + 1}) d t = - \frac{π}{4} + \frac{π}{4} - \int_{0}^{1} \frac{t^{2} - 2}{t^{4} - t^{2} + 1} d t

= - \frac{1}{2} \int_{0}^{1} \frac{3 (t^{2} - 1) - (t^{2} + 1)}{t^{4} - t^{2} + 1} d t = \frac{1}{2} \int_{0}^{1} \frac{t^{2} + 1}{t^{4} - t^{2} + 1} d t - \frac{3}{2} \int_{0}^{1} \frac{t^{2} - 1}{t^{4} - t^{2} + 1} d t

= \frac{1}{2} \int_{0}^{1} \frac{1 + \frac{1}{t^{2}}}{{(t - \frac{1}{t})}^{2} + 1} d t - \frac{3}{2} \int_{0}^{1} \frac{1 - \frac{1}{t^{2}}}{{(t + \frac{1}{t})}^{2} - 3} d t

= \frac{1}{2} {[\arctan (\frac{t^{2} - 1}{t})]}_{0}^{1} + \frac{\sqrt{3}}{4} {[\ln ∣ \frac{t^{2} + \sqrt{3} t + 1}{t^{2} - \sqrt{3} t + 1} ∣]}_{0}^{1}

= \frac{1}{2} (\frac{π}{2}) + \frac{\sqrt{3}}{4} \ln ∣ \frac{2 + \sqrt{3}}{2 - \sqrt{3}} ∣= \frac{π}{4} + \frac{\sqrt{3}}{4} \ln {(2 + \sqrt{3})}^{2}

= \frac{π}{4} + \frac{\sqrt{3}}{4} \ln (7 + 4 \sqrt{3})

Commented by mnjuly1970 last updated on 31/Jan/22

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