How to integrate $tan^{-1}left(frac{1}{2 sin(x)}right)$?












3












$begingroup$


I want to calculate the following integral :



$displaystyle{int^{frac{pi}{2}}_{0} tan^{-1}left(frac{1}{2 sin(x)}right)} text{ d}x$



But I don't how; I tried by subsituting $u = frac{1}{2 sin(x)}$ and $u = tan^{-1}left(frac{1}{2 sin(x)}right)$ but it doesn't lead me anywhere.



Thanks for your help.






calculus integration trigonometry definite-integrals trigonometric-integrals







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    3












    $begingroup$


    I want to calculate the following integral :



    $displaystyle{int^{frac{pi}{2}}_{0} tan^{-1}left(frac{1}{2 sin(x)}right)} text{ d}x$



    But I don't how; I tried by subsituting $u = frac{1}{2 sin(x)}$ and $u = tan^{-1}left(frac{1}{2 sin(x)}right)$ but it doesn't lead me anywhere.



    Thanks for your help.






    calculus integration trigonometry definite-integrals trigonometric-integrals







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      I want to calculate the following integral :



      $displaystyle{int^{frac{pi}{2}}_{0} tan^{-1}left(frac{1}{2 sin(x)}right)} text{ d}x$



      But I don't how; I tried by subsituting $u = frac{1}{2 sin(x)}$ and $u = tan^{-1}left(frac{1}{2 sin(x)}right)$ but it doesn't lead me anywhere.



      Thanks for your help.






      calculus integration trigonometry definite-integrals trigonometric-integrals







      share|cite|improve this question











      $endgroup$




      I want to calculate the following integral :



      $displaystyle{int^{frac{pi}{2}}_{0} tan^{-1}left(frac{1}{2 sin(x)}right)} text{ d}x$



      But I don't how; I tried by subsituting $u = frac{1}{2 sin(x)}$ and $u = tan^{-1}left(frac{1}{2 sin(x)}right)$ but it doesn't lead me anywhere.



      Thanks for your help.






      calculus integration trigonometry definite-integrals trigonometric-integrals




      calculus integration trigonometry definite-integrals trigonometric-integrals






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      edited Dec 9 '18 at 12:10









      DavidG

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      asked Nov 1 '18 at 21:19









      SkyostSkyost

      184




      184






















          2 Answers
          2






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          1












          $begingroup$

          Well, the integral over $(0,2pi)$ is clearly zero by symmetry, hence the given problem is equivalent to finding



          $$ -int_{0}^{pi/2}arctanleft(frac{1}{2sin x}right),dx=-int_{0}^{1}frac{arctanfrac{1}{2x}}{sqrt{1-x^2}},dx=-frac{pi^2}{4}+int_{0}^{1}frac{arctan(2x)}{sqrt{1-x^2}},dx $$
          or
          $$ -frac{pi^2}{4}+int_{0}^{2}int_{0}^{1}frac{x}{(1+a^2 x^2)sqrt{1-x^2}},dx,da=-frac{pi^2}{4}+int_{0}^{2}frac{text{arcsinh}(a)}{asqrt{1+a^2}},da$$
          or
          $$ -frac{pi^2}{4}+int_{0}^{log(2+sqrt{5})}frac{u}{sinh u},du =-frac{pi^2}{4}+int_{1}^{2+sqrt{5}}frac{2log v}{v^2-1},dv$$
          where the last integral depends on the dilogarithm $text{Li}_2$ evaluated at $pm(sqrt{5}-2)$:




          $$ int_{0}^{pi/2}arctanleft(frac{1}{2sin x}right),dx= text{arcsinh}left(tfrac{1}{2}right)text{arcsinh}(2)-text{Li}_2(sqrt{5}-2)+text{Li}_2(2-sqrt{5})$$







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you a lot ! 😉
            $endgroup$
            – Skyost
            Nov 2 '18 at 11:43



















          0












          $begingroup$

          Partial answer: Consider $f(y) = int_{pi/2}^{2pi} arctanfrac1{ysin x},dx$. You are interested ultimately in $f(2)$. It appears that $f'(y)$ can be integrated/evaluated (pass the derivative through the integral, simplify, and do $u=cos x$) easily enough. Can you then integrate this result to get back $f(y)$?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Simpler to say it than actually doing it, isn't it?
            $endgroup$
            – Jack D'Aurizio
            Nov 2 '18 at 3:21











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Well, the integral over $(0,2pi)$ is clearly zero by symmetry, hence the given problem is equivalent to finding



          $$ -int_{0}^{pi/2}arctanleft(frac{1}{2sin x}right),dx=-int_{0}^{1}frac{arctanfrac{1}{2x}}{sqrt{1-x^2}},dx=-frac{pi^2}{4}+int_{0}^{1}frac{arctan(2x)}{sqrt{1-x^2}},dx $$
          or
          $$ -frac{pi^2}{4}+int_{0}^{2}int_{0}^{1}frac{x}{(1+a^2 x^2)sqrt{1-x^2}},dx,da=-frac{pi^2}{4}+int_{0}^{2}frac{text{arcsinh}(a)}{asqrt{1+a^2}},da$$
          or
          $$ -frac{pi^2}{4}+int_{0}^{log(2+sqrt{5})}frac{u}{sinh u},du =-frac{pi^2}{4}+int_{1}^{2+sqrt{5}}frac{2log v}{v^2-1},dv$$
          where the last integral depends on the dilogarithm $text{Li}_2$ evaluated at $pm(sqrt{5}-2)$:




          $$ int_{0}^{pi/2}arctanleft(frac{1}{2sin x}right),dx= text{arcsinh}left(tfrac{1}{2}right)text{arcsinh}(2)-text{Li}_2(sqrt{5}-2)+text{Li}_2(2-sqrt{5})$$







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you a lot ! 😉
            $endgroup$
            – Skyost
            Nov 2 '18 at 11:43
















          1












          $begingroup$

          Well, the integral over $(0,2pi)$ is clearly zero by symmetry, hence the given problem is equivalent to finding



          $$ -int_{0}^{pi/2}arctanleft(frac{1}{2sin x}right),dx=-int_{0}^{1}frac{arctanfrac{1}{2x}}{sqrt{1-x^2}},dx=-frac{pi^2}{4}+int_{0}^{1}frac{arctan(2x)}{sqrt{1-x^2}},dx $$
          or
          $$ -frac{pi^2}{4}+int_{0}^{2}int_{0}^{1}frac{x}{(1+a^2 x^2)sqrt{1-x^2}},dx,da=-frac{pi^2}{4}+int_{0}^{2}frac{text{arcsinh}(a)}{asqrt{1+a^2}},da$$
          or
          $$ -frac{pi^2}{4}+int_{0}^{log(2+sqrt{5})}frac{u}{sinh u},du =-frac{pi^2}{4}+int_{1}^{2+sqrt{5}}frac{2log v}{v^2-1},dv$$
          where the last integral depends on the dilogarithm $text{Li}_2$ evaluated at $pm(sqrt{5}-2)$:




          $$ int_{0}^{pi/2}arctanleft(frac{1}{2sin x}right),dx= text{arcsinh}left(tfrac{1}{2}right)text{arcsinh}(2)-text{Li}_2(sqrt{5}-2)+text{Li}_2(2-sqrt{5})$$







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you a lot ! 😉
            $endgroup$
            – Skyost
            Nov 2 '18 at 11:43














          1












          1








          1





          $begingroup$

          Well, the integral over $(0,2pi)$ is clearly zero by symmetry, hence the given problem is equivalent to finding



          $$ -int_{0}^{pi/2}arctanleft(frac{1}{2sin x}right),dx=-int_{0}^{1}frac{arctanfrac{1}{2x}}{sqrt{1-x^2}},dx=-frac{pi^2}{4}+int_{0}^{1}frac{arctan(2x)}{sqrt{1-x^2}},dx $$
          or
          $$ -frac{pi^2}{4}+int_{0}^{2}int_{0}^{1}frac{x}{(1+a^2 x^2)sqrt{1-x^2}},dx,da=-frac{pi^2}{4}+int_{0}^{2}frac{text{arcsinh}(a)}{asqrt{1+a^2}},da$$
          or
          $$ -frac{pi^2}{4}+int_{0}^{log(2+sqrt{5})}frac{u}{sinh u},du =-frac{pi^2}{4}+int_{1}^{2+sqrt{5}}frac{2log v}{v^2-1},dv$$
          where the last integral depends on the dilogarithm $text{Li}_2$ evaluated at $pm(sqrt{5}-2)$:




          $$ int_{0}^{pi/2}arctanleft(frac{1}{2sin x}right),dx= text{arcsinh}left(tfrac{1}{2}right)text{arcsinh}(2)-text{Li}_2(sqrt{5}-2)+text{Li}_2(2-sqrt{5})$$







          share|cite|improve this answer









          $endgroup$



          Well, the integral over $(0,2pi)$ is clearly zero by symmetry, hence the given problem is equivalent to finding



          $$ -int_{0}^{pi/2}arctanleft(frac{1}{2sin x}right),dx=-int_{0}^{1}frac{arctanfrac{1}{2x}}{sqrt{1-x^2}},dx=-frac{pi^2}{4}+int_{0}^{1}frac{arctan(2x)}{sqrt{1-x^2}},dx $$
          or
          $$ -frac{pi^2}{4}+int_{0}^{2}int_{0}^{1}frac{x}{(1+a^2 x^2)sqrt{1-x^2}},dx,da=-frac{pi^2}{4}+int_{0}^{2}frac{text{arcsinh}(a)}{asqrt{1+a^2}},da$$
          or
          $$ -frac{pi^2}{4}+int_{0}^{log(2+sqrt{5})}frac{u}{sinh u},du =-frac{pi^2}{4}+int_{1}^{2+sqrt{5}}frac{2log v}{v^2-1},dv$$
          where the last integral depends on the dilogarithm $text{Li}_2$ evaluated at $pm(sqrt{5}-2)$:




          $$ int_{0}^{pi/2}arctanleft(frac{1}{2sin x}right),dx= text{arcsinh}left(tfrac{1}{2}right)text{arcsinh}(2)-text{Li}_2(sqrt{5}-2)+text{Li}_2(2-sqrt{5})$$








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          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 2 '18 at 3:17









          Jack D'AurizioJack D'Aurizio

          291k33284669




          291k33284669












          • $begingroup$
            Thank you a lot ! 😉
            $endgroup$
            – Skyost
            Nov 2 '18 at 11:43


















          • $begingroup$
            Thank you a lot ! 😉
            $endgroup$
            – Skyost
            Nov 2 '18 at 11:43
















          $begingroup$
          Thank you a lot ! 😉
          $endgroup$
          – Skyost
          Nov 2 '18 at 11:43




          $begingroup$
          Thank you a lot ! 😉
          $endgroup$
          – Skyost
          Nov 2 '18 at 11:43











          0












          $begingroup$

          Partial answer: Consider $f(y) = int_{pi/2}^{2pi} arctanfrac1{ysin x},dx$. You are interested ultimately in $f(2)$. It appears that $f'(y)$ can be integrated/evaluated (pass the derivative through the integral, simplify, and do $u=cos x$) easily enough. Can you then integrate this result to get back $f(y)$?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Simpler to say it than actually doing it, isn't it?
            $endgroup$
            – Jack D'Aurizio
            Nov 2 '18 at 3:21
















          0












          $begingroup$

          Partial answer: Consider $f(y) = int_{pi/2}^{2pi} arctanfrac1{ysin x},dx$. You are interested ultimately in $f(2)$. It appears that $f'(y)$ can be integrated/evaluated (pass the derivative through the integral, simplify, and do $u=cos x$) easily enough. Can you then integrate this result to get back $f(y)$?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Simpler to say it than actually doing it, isn't it?
            $endgroup$
            – Jack D'Aurizio
            Nov 2 '18 at 3:21














          0












          0








          0





          $begingroup$

          Partial answer: Consider $f(y) = int_{pi/2}^{2pi} arctanfrac1{ysin x},dx$. You are interested ultimately in $f(2)$. It appears that $f'(y)$ can be integrated/evaluated (pass the derivative through the integral, simplify, and do $u=cos x$) easily enough. Can you then integrate this result to get back $f(y)$?






          share|cite|improve this answer









          $endgroup$



          Partial answer: Consider $f(y) = int_{pi/2}^{2pi} arctanfrac1{ysin x},dx$. You are interested ultimately in $f(2)$. It appears that $f'(y)$ can be integrated/evaluated (pass the derivative through the integral, simplify, and do $u=cos x$) easily enough. Can you then integrate this result to get back $f(y)$?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 1 '18 at 22:00









          JasonJason

          1,367610




          1,367610












          • $begingroup$
            Simpler to say it than actually doing it, isn't it?
            $endgroup$
            – Jack D'Aurizio
            Nov 2 '18 at 3:21


















          • $begingroup$
            Simpler to say it than actually doing it, isn't it?
            $endgroup$
            – Jack D'Aurizio
            Nov 2 '18 at 3:21
















          $begingroup$
          Simpler to say it than actually doing it, isn't it?
          $endgroup$
          – Jack D'Aurizio
          Nov 2 '18 at 3:21




          $begingroup$
          Simpler to say it than actually doing it, isn't it?
          $endgroup$
          – Jack D'Aurizio
          Nov 2 '18 at 3:21


















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