I-0-pi-2-tan-1-sinx-2-dx-

Question Number 199570 by universe last updated on 05/Nov/23

I = \int_{0}^{π / 2} \tan^{- 1} (\frac{\sin x}{2}) d x

Answered by witcher3 last updated on 05/Nov/23

I (a) = \int_{0}^{\frac{π}{2}} \tan^{- 1} (asin (x)) dx

I = I (\frac{1}{2})

I (0) = 0

I^{'} (a) = \int_{0}^{\frac{π}{2}} \frac{\sin (x)}{1 + a^{2} \sin^{2} (x)} dx

= \int_{0}^{1} \frac{dy}{1 + a^{2} (1 - y^{2})} dy

= \int_{0}^{1} \frac{dy}{[\sqrt{1 + a^{2}} - ay] [\sqrt{a^{2} + 1} + ay]}

= \frac{1}{2 \sqrt{a^{2} + 1}} \int_{0}^{1} {\frac{1}{\sqrt{1 + a^{2}} - ay} + \frac{1}{ay + \sqrt{1 + a^{2}}}} dy

= \frac{1}{2 a \sqrt{1 + a^{2}}} \ln (\frac{a + \sqrt{1 + a^{2}}}{\sqrt{1 + a^{2}} - a}) = I^{'} (a)

I (\frac{1}{2}) = \int_{0}^{\frac{1}{2}} \frac{1}{2 a \sqrt{1 + a^{2}}} \ln (\frac{a + \sqrt{1 + a^{2}}}{\sqrt{1 + a^{2}} - a}) da

a = sh (x) \Rightarrow da = ch (x) dx

I = \int_{0}^{{sh}^{-} (\frac{1}{2})} \frac{1}{2 sh (x)} \ln (\frac{e^{x}}{e^{- x}}) = \int_{0}^{{sh}^{-} (\frac{1}{2})} \frac{x}{e^{x} - e^{- x}} dx

= \frac{1}{2} \int_{0}^{{sh}^{-} (\frac{1}{2})} \frac{x}{e^{x} - 1} + \frac{x}{e^{x} + 1} dx = I

\int \frac{x}{e^{x} - 1} = xln (1 - e^{- x}) - \int \frac{\ln (1 - e^{- x}) {de}^{- x}}{e^{- x}}

= xln (1 - e^{- x}) - {Li}_{2} (e^{- x})

\int \frac{x}{e^{x} + 1} dx = - xln (e^{- x} + 1) + {Li}_{2} (- e^{- x})

I = \frac{1}{2} ({sh}^{-} (\frac{1}{2}) \ln (1 - e^{- {sh}^{-} (\frac{1}{2})}) - {Li}_{2} (e^{- {sh}^{-} (\frac{1}{2})})

- {sh}^{-} (\frac{1}{2}) \ln (1 + e^{- {sh}^{-} (\frac{1}{2})}) + {Li}_{2} (- e^{- {sh}^{-} (\frac{1}{2})}) + {Li}_{2} (1) - L i_{2} (- 1)

= \frac{1}{12} (π^{2} - {6 \sinh}^{-} (2) {csch}^{-} (2))

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