Evaluating $int_0^2(tan^{-1}(pi x)-tan^{-1} x),mathrm{d}x$












6












$begingroup$


Hint given: Write the integrand as an integral.



I'm supposed to do this as double integration.



My attempt:



$$int_0^2 [tan^{-1}y]^{pi x}_{x}$$



$$= int_0^2 int_x^{pi x} frac { mathrm{d}y mathrm{d}x} {y^2+1}$$



$$= int_2^{2pi} int_{frac{y}{pi}}^2 frac { mathrm{d}x mathrm{d}y} {y^2+1}$$



$$= int_2^{2pi} frac { [x]^2_{frac{y}{pi}} mathrm{d}y } { y^2+1}$$



$$= int_2^{2pi} frac { 2- {frac{y}{pi}} mathrm{d}y } { y^2+1}$$



Carrying out this integration, I got, $$2[tan^{-1} 2 pi - tan^{-1} 2] - frac {1}{2 pi} [ln frac {1+4 {pi}^2} {5}]$$



But I'm supposed to get $$2[tan^{-1} 2 pi - tan^{-1} 2] - frac {1}{2 pi} [ln frac {1+4 {pi}^2} {5}]+ [frac {pi-1}{2 pi}] ln 5$$



Can someone please explain where I'm wrong? I've failed to pinpoint my mistake. Thank you.






integration







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  • 2




    $begingroup$
    At third line it seems to me the integral must be $$int_0^{2pi}int_{frac{y}{pi}}^{y}frac{operatorname d!x operatorname d!y}{y^2+1}$$
    $endgroup$
    – Ángel Mario Gallegos
    Nov 4 '14 at 6:13










  • $begingroup$
    but that is taking my answer even further away from what I require.
    $endgroup$
    – Diya
    Nov 4 '14 at 6:50










  • $begingroup$
    I got it! You were right! The integral is divided into two regions. One is the one I used, and the other is the one you mentioned. Thank you so much! :)
    $endgroup$
    – Diya
    Nov 4 '14 at 6:57
















6












$begingroup$


Hint given: Write the integrand as an integral.



I'm supposed to do this as double integration.



My attempt:



$$int_0^2 [tan^{-1}y]^{pi x}_{x}$$



$$= int_0^2 int_x^{pi x} frac { mathrm{d}y mathrm{d}x} {y^2+1}$$



$$= int_2^{2pi} int_{frac{y}{pi}}^2 frac { mathrm{d}x mathrm{d}y} {y^2+1}$$



$$= int_2^{2pi} frac { [x]^2_{frac{y}{pi}} mathrm{d}y } { y^2+1}$$



$$= int_2^{2pi} frac { 2- {frac{y}{pi}} mathrm{d}y } { y^2+1}$$



Carrying out this integration, I got, $$2[tan^{-1} 2 pi - tan^{-1} 2] - frac {1}{2 pi} [ln frac {1+4 {pi}^2} {5}]$$



But I'm supposed to get $$2[tan^{-1} 2 pi - tan^{-1} 2] - frac {1}{2 pi} [ln frac {1+4 {pi}^2} {5}]+ [frac {pi-1}{2 pi}] ln 5$$



Can someone please explain where I'm wrong? I've failed to pinpoint my mistake. Thank you.






integration







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    At third line it seems to me the integral must be $$int_0^{2pi}int_{frac{y}{pi}}^{y}frac{operatorname d!x operatorname d!y}{y^2+1}$$
    $endgroup$
    – Ángel Mario Gallegos
    Nov 4 '14 at 6:13










  • $begingroup$
    but that is taking my answer even further away from what I require.
    $endgroup$
    – Diya
    Nov 4 '14 at 6:50










  • $begingroup$
    I got it! You were right! The integral is divided into two regions. One is the one I used, and the other is the one you mentioned. Thank you so much! :)
    $endgroup$
    – Diya
    Nov 4 '14 at 6:57














6












6








6


1



$begingroup$


Hint given: Write the integrand as an integral.



I'm supposed to do this as double integration.



My attempt:



$$int_0^2 [tan^{-1}y]^{pi x}_{x}$$



$$= int_0^2 int_x^{pi x} frac { mathrm{d}y mathrm{d}x} {y^2+1}$$



$$= int_2^{2pi} int_{frac{y}{pi}}^2 frac { mathrm{d}x mathrm{d}y} {y^2+1}$$



$$= int_2^{2pi} frac { [x]^2_{frac{y}{pi}} mathrm{d}y } { y^2+1}$$



$$= int_2^{2pi} frac { 2- {frac{y}{pi}} mathrm{d}y } { y^2+1}$$



Carrying out this integration, I got, $$2[tan^{-1} 2 pi - tan^{-1} 2] - frac {1}{2 pi} [ln frac {1+4 {pi}^2} {5}]$$



But I'm supposed to get $$2[tan^{-1} 2 pi - tan^{-1} 2] - frac {1}{2 pi} [ln frac {1+4 {pi}^2} {5}]+ [frac {pi-1}{2 pi}] ln 5$$



Can someone please explain where I'm wrong? I've failed to pinpoint my mistake. Thank you.






integration







share|cite|improve this question











$endgroup$




Hint given: Write the integrand as an integral.



I'm supposed to do this as double integration.



My attempt:



$$int_0^2 [tan^{-1}y]^{pi x}_{x}$$



$$= int_0^2 int_x^{pi x} frac { mathrm{d}y mathrm{d}x} {y^2+1}$$



$$= int_2^{2pi} int_{frac{y}{pi}}^2 frac { mathrm{d}x mathrm{d}y} {y^2+1}$$



$$= int_2^{2pi} frac { [x]^2_{frac{y}{pi}} mathrm{d}y } { y^2+1}$$



$$= int_2^{2pi} frac { 2- {frac{y}{pi}} mathrm{d}y } { y^2+1}$$



Carrying out this integration, I got, $$2[tan^{-1} 2 pi - tan^{-1} 2] - frac {1}{2 pi} [ln frac {1+4 {pi}^2} {5}]$$



But I'm supposed to get $$2[tan^{-1} 2 pi - tan^{-1} 2] - frac {1}{2 pi} [ln frac {1+4 {pi}^2} {5}]+ [frac {pi-1}{2 pi}] ln 5$$



Can someone please explain where I'm wrong? I've failed to pinpoint my mistake. Thank you.






integration




integration






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edited Dec 25 '18 at 5:26









Saad

19.7k92352




19.7k92352










asked Nov 4 '14 at 6:00









DiyaDiya

1,2862925




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  • 2




    $begingroup$
    At third line it seems to me the integral must be $$int_0^{2pi}int_{frac{y}{pi}}^{y}frac{operatorname d!x operatorname d!y}{y^2+1}$$
    $endgroup$
    – Ángel Mario Gallegos
    Nov 4 '14 at 6:13










  • $begingroup$
    but that is taking my answer even further away from what I require.
    $endgroup$
    – Diya
    Nov 4 '14 at 6:50










  • $begingroup$
    I got it! You were right! The integral is divided into two regions. One is the one I used, and the other is the one you mentioned. Thank you so much! :)
    $endgroup$
    – Diya
    Nov 4 '14 at 6:57














  • 2




    $begingroup$
    At third line it seems to me the integral must be $$int_0^{2pi}int_{frac{y}{pi}}^{y}frac{operatorname d!x operatorname d!y}{y^2+1}$$
    $endgroup$
    – Ángel Mario Gallegos
    Nov 4 '14 at 6:13










  • $begingroup$
    but that is taking my answer even further away from what I require.
    $endgroup$
    – Diya
    Nov 4 '14 at 6:50










  • $begingroup$
    I got it! You were right! The integral is divided into two regions. One is the one I used, and the other is the one you mentioned. Thank you so much! :)
    $endgroup$
    – Diya
    Nov 4 '14 at 6:57








2




2




$begingroup$
At third line it seems to me the integral must be $$int_0^{2pi}int_{frac{y}{pi}}^{y}frac{operatorname d!x operatorname d!y}{y^2+1}$$
$endgroup$
– Ángel Mario Gallegos
Nov 4 '14 at 6:13




$begingroup$
At third line it seems to me the integral must be $$int_0^{2pi}int_{frac{y}{pi}}^{y}frac{operatorname d!x operatorname d!y}{y^2+1}$$
$endgroup$
– Ángel Mario Gallegos
Nov 4 '14 at 6:13












$begingroup$
but that is taking my answer even further away from what I require.
$endgroup$
– Diya
Nov 4 '14 at 6:50




$begingroup$
but that is taking my answer even further away from what I require.
$endgroup$
– Diya
Nov 4 '14 at 6:50












$begingroup$
I got it! You were right! The integral is divided into two regions. One is the one I used, and the other is the one you mentioned. Thank you so much! :)
$endgroup$
– Diya
Nov 4 '14 at 6:57




$begingroup$
I got it! You were right! The integral is divided into two regions. One is the one I used, and the other is the one you mentioned. Thank you so much! :)
$endgroup$
– Diya
Nov 4 '14 at 6:57










3 Answers
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$begingroup$

Use the change of variables $y = xt.$



$$I = int_0^2 int_x^{pi x} frac { mathrm{d}y , mathrm{d}x} {y^2+1}\=int_0^2 int_1^{pi } frac { x} {x^2t^2+1}mathrm{d}t , mathrm{d}x\=int_1^{pi} int_0^{2 } frac { x} {x^2t^2+1}mathrm{d}x , mathrm{d}t\=int_1^{pi} frac{ln(1+4t^2)}{2t^2} mathrm{d}t$$



Now use integration by parts.



$$I = -left.frac{ln(1+4t^2)}{2t}right|_1^{pi}+4int_1^{pi}frac1{1+4t^2} , dt$$






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    3












    $begingroup$

    A more straight forward approach uses integration by parts.



    Define:
    begin{align}
    & I(c)=int_{a}^{b}dx(1 times arctan{c x})=int_{ac}^{bc}frac{dy}{c} (1 timesarctan{ y})=\&frac{1}{c}y arctan(y)|_{ac}^{bc}-frac{1}{2c}int_{ac}^{bc}frac{y}{1+y^2}
    end{align}



    using partial fraction this reads:
    begin{align}
    I(c)=frac{1}{c}y arctan(y)|_{ac}^{bc}-frac{1}{2c}log(1+y^2)|_{ac}^{bc}
    end{align}



    taking $I(pi)-I(0)$ with $a=0$ and $b=2$ we are done






    share|cite|improve this answer











    $endgroup$





















      2












      $begingroup$

      For the sake of an alternative approach, recall the formula for the integral of an inverse function
      $$int f^{-1}(x)mathrm{d}x=xf^{-1}(x)-Fcirc f^{-1}(x)$$
      Where $F'(x)=f(x)$. Plugging in $f(x)=tan(x)$,
      $$I=intarctan(x)mathrm{d}x=xarctan(x)-int_0^{arctan(x)}tan(t)mathrm{d}t$$
      Then recall that $$(-log|cos(x)|)'=tan(x)$$
      So we have
      $$I=xarctan(x)+log|cos(arctan(x))|$$
      then using trig,
      $$I=xarctan(x)-frac12log(x^2+1)$$
      So
      $$I_1=int_0^2arctan(x)mathrm{d}x=2arctan2-frac12log5$$
      And
      $$
      begin{align}
      I_2=&int_0^2arctan(pi x)mathrm{d}x\
      =&frac1piint_0^{2pi}arctan(x)mathrm{d}x\
      =&frac1pibigg(2piarctan2pi-frac12log(4pi^2+1)bigg)\
      =&2arctan2pi-frac1{2pi}log(4pi^2+1)\
      end{align}
      $$
      So
      $$
      begin{align}
      int_0^2(arctanpi x-arctan x)mathrm{d}x=&I_2-I_1\
      =&2arctan2pi-frac1{2pi}log(4pi^2+1)-2arctan2+frac12log5\
      =&2arctan2pi-frac1{2}logsqrt[pi]{4pi^2+1}-2arctan2+frac12log5\
      =&2(arctan2pi-arctan2)+frac12logfrac{5}{sqrt[pi]{4pi^2+1}}\
      end{align}
      $$






      share|cite|improve this answer









      $endgroup$













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        3 Answers
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        3 Answers
        3






        active

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        active

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        active

        oldest

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        3












        $begingroup$

        Use the change of variables $y = xt.$



        $$I = int_0^2 int_x^{pi x} frac { mathrm{d}y , mathrm{d}x} {y^2+1}\=int_0^2 int_1^{pi } frac { x} {x^2t^2+1}mathrm{d}t , mathrm{d}x\=int_1^{pi} int_0^{2 } frac { x} {x^2t^2+1}mathrm{d}x , mathrm{d}t\=int_1^{pi} frac{ln(1+4t^2)}{2t^2} mathrm{d}t$$



        Now use integration by parts.



        $$I = -left.frac{ln(1+4t^2)}{2t}right|_1^{pi}+4int_1^{pi}frac1{1+4t^2} , dt$$






        share|cite|improve this answer











        $endgroup$


















          3












          $begingroup$

          Use the change of variables $y = xt.$



          $$I = int_0^2 int_x^{pi x} frac { mathrm{d}y , mathrm{d}x} {y^2+1}\=int_0^2 int_1^{pi } frac { x} {x^2t^2+1}mathrm{d}t , mathrm{d}x\=int_1^{pi} int_0^{2 } frac { x} {x^2t^2+1}mathrm{d}x , mathrm{d}t\=int_1^{pi} frac{ln(1+4t^2)}{2t^2} mathrm{d}t$$



          Now use integration by parts.



          $$I = -left.frac{ln(1+4t^2)}{2t}right|_1^{pi}+4int_1^{pi}frac1{1+4t^2} , dt$$






          share|cite|improve this answer











          $endgroup$
















            3












            3








            3





            $begingroup$

            Use the change of variables $y = xt.$



            $$I = int_0^2 int_x^{pi x} frac { mathrm{d}y , mathrm{d}x} {y^2+1}\=int_0^2 int_1^{pi } frac { x} {x^2t^2+1}mathrm{d}t , mathrm{d}x\=int_1^{pi} int_0^{2 } frac { x} {x^2t^2+1}mathrm{d}x , mathrm{d}t\=int_1^{pi} frac{ln(1+4t^2)}{2t^2} mathrm{d}t$$



            Now use integration by parts.



            $$I = -left.frac{ln(1+4t^2)}{2t}right|_1^{pi}+4int_1^{pi}frac1{1+4t^2} , dt$$






            share|cite|improve this answer











            $endgroup$



            Use the change of variables $y = xt.$



            $$I = int_0^2 int_x^{pi x} frac { mathrm{d}y , mathrm{d}x} {y^2+1}\=int_0^2 int_1^{pi } frac { x} {x^2t^2+1}mathrm{d}t , mathrm{d}x\=int_1^{pi} int_0^{2 } frac { x} {x^2t^2+1}mathrm{d}x , mathrm{d}t\=int_1^{pi} frac{ln(1+4t^2)}{2t^2} mathrm{d}t$$



            Now use integration by parts.



            $$I = -left.frac{ln(1+4t^2)}{2t}right|_1^{pi}+4int_1^{pi}frac1{1+4t^2} , dt$$







            share|cite|improve this answer














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            share|cite|improve this answer








            edited Nov 4 '14 at 6:48

























            answered Nov 4 '14 at 6:30









            RRLRRL

            50.9k42573




            50.9k42573























                3












                $begingroup$

                A more straight forward approach uses integration by parts.



                Define:
                begin{align}
                & I(c)=int_{a}^{b}dx(1 times arctan{c x})=int_{ac}^{bc}frac{dy}{c} (1 timesarctan{ y})=\&frac{1}{c}y arctan(y)|_{ac}^{bc}-frac{1}{2c}int_{ac}^{bc}frac{y}{1+y^2}
                end{align}



                using partial fraction this reads:
                begin{align}
                I(c)=frac{1}{c}y arctan(y)|_{ac}^{bc}-frac{1}{2c}log(1+y^2)|_{ac}^{bc}
                end{align}



                taking $I(pi)-I(0)$ with $a=0$ and $b=2$ we are done






                share|cite|improve this answer











                $endgroup$


















                  3












                  $begingroup$

                  A more straight forward approach uses integration by parts.



                  Define:
                  begin{align}
                  & I(c)=int_{a}^{b}dx(1 times arctan{c x})=int_{ac}^{bc}frac{dy}{c} (1 timesarctan{ y})=\&frac{1}{c}y arctan(y)|_{ac}^{bc}-frac{1}{2c}int_{ac}^{bc}frac{y}{1+y^2}
                  end{align}



                  using partial fraction this reads:
                  begin{align}
                  I(c)=frac{1}{c}y arctan(y)|_{ac}^{bc}-frac{1}{2c}log(1+y^2)|_{ac}^{bc}
                  end{align}



                  taking $I(pi)-I(0)$ with $a=0$ and $b=2$ we are done






                  share|cite|improve this answer











                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    A more straight forward approach uses integration by parts.



                    Define:
                    begin{align}
                    & I(c)=int_{a}^{b}dx(1 times arctan{c x})=int_{ac}^{bc}frac{dy}{c} (1 timesarctan{ y})=\&frac{1}{c}y arctan(y)|_{ac}^{bc}-frac{1}{2c}int_{ac}^{bc}frac{y}{1+y^2}
                    end{align}



                    using partial fraction this reads:
                    begin{align}
                    I(c)=frac{1}{c}y arctan(y)|_{ac}^{bc}-frac{1}{2c}log(1+y^2)|_{ac}^{bc}
                    end{align}



                    taking $I(pi)-I(0)$ with $a=0$ and $b=2$ we are done






                    share|cite|improve this answer











                    $endgroup$



                    A more straight forward approach uses integration by parts.



                    Define:
                    begin{align}
                    & I(c)=int_{a}^{b}dx(1 times arctan{c x})=int_{ac}^{bc}frac{dy}{c} (1 timesarctan{ y})=\&frac{1}{c}y arctan(y)|_{ac}^{bc}-frac{1}{2c}int_{ac}^{bc}frac{y}{1+y^2}
                    end{align}



                    using partial fraction this reads:
                    begin{align}
                    I(c)=frac{1}{c}y arctan(y)|_{ac}^{bc}-frac{1}{2c}log(1+y^2)|_{ac}^{bc}
                    end{align}



                    taking $I(pi)-I(0)$ with $a=0$ and $b=2$ we are done







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 4 '14 at 17:59

























                    answered Nov 4 '14 at 17:20









                    tiredtired

                    10.6k12043




                    10.6k12043























                        2












                        $begingroup$

                        For the sake of an alternative approach, recall the formula for the integral of an inverse function
                        $$int f^{-1}(x)mathrm{d}x=xf^{-1}(x)-Fcirc f^{-1}(x)$$
                        Where $F'(x)=f(x)$. Plugging in $f(x)=tan(x)$,
                        $$I=intarctan(x)mathrm{d}x=xarctan(x)-int_0^{arctan(x)}tan(t)mathrm{d}t$$
                        Then recall that $$(-log|cos(x)|)'=tan(x)$$
                        So we have
                        $$I=xarctan(x)+log|cos(arctan(x))|$$
                        then using trig,
                        $$I=xarctan(x)-frac12log(x^2+1)$$
                        So
                        $$I_1=int_0^2arctan(x)mathrm{d}x=2arctan2-frac12log5$$
                        And
                        $$
                        begin{align}
                        I_2=&int_0^2arctan(pi x)mathrm{d}x\
                        =&frac1piint_0^{2pi}arctan(x)mathrm{d}x\
                        =&frac1pibigg(2piarctan2pi-frac12log(4pi^2+1)bigg)\
                        =&2arctan2pi-frac1{2pi}log(4pi^2+1)\
                        end{align}
                        $$
                        So
                        $$
                        begin{align}
                        int_0^2(arctanpi x-arctan x)mathrm{d}x=&I_2-I_1\
                        =&2arctan2pi-frac1{2pi}log(4pi^2+1)-2arctan2+frac12log5\
                        =&2arctan2pi-frac1{2}logsqrt[pi]{4pi^2+1}-2arctan2+frac12log5\
                        =&2(arctan2pi-arctan2)+frac12logfrac{5}{sqrt[pi]{4pi^2+1}}\
                        end{align}
                        $$






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          For the sake of an alternative approach, recall the formula for the integral of an inverse function
                          $$int f^{-1}(x)mathrm{d}x=xf^{-1}(x)-Fcirc f^{-1}(x)$$
                          Where $F'(x)=f(x)$. Plugging in $f(x)=tan(x)$,
                          $$I=intarctan(x)mathrm{d}x=xarctan(x)-int_0^{arctan(x)}tan(t)mathrm{d}t$$
                          Then recall that $$(-log|cos(x)|)'=tan(x)$$
                          So we have
                          $$I=xarctan(x)+log|cos(arctan(x))|$$
                          then using trig,
                          $$I=xarctan(x)-frac12log(x^2+1)$$
                          So
                          $$I_1=int_0^2arctan(x)mathrm{d}x=2arctan2-frac12log5$$
                          And
                          $$
                          begin{align}
                          I_2=&int_0^2arctan(pi x)mathrm{d}x\
                          =&frac1piint_0^{2pi}arctan(x)mathrm{d}x\
                          =&frac1pibigg(2piarctan2pi-frac12log(4pi^2+1)bigg)\
                          =&2arctan2pi-frac1{2pi}log(4pi^2+1)\
                          end{align}
                          $$
                          So
                          $$
                          begin{align}
                          int_0^2(arctanpi x-arctan x)mathrm{d}x=&I_2-I_1\
                          =&2arctan2pi-frac1{2pi}log(4pi^2+1)-2arctan2+frac12log5\
                          =&2arctan2pi-frac1{2}logsqrt[pi]{4pi^2+1}-2arctan2+frac12log5\
                          =&2(arctan2pi-arctan2)+frac12logfrac{5}{sqrt[pi]{4pi^2+1}}\
                          end{align}
                          $$






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            For the sake of an alternative approach, recall the formula for the integral of an inverse function
                            $$int f^{-1}(x)mathrm{d}x=xf^{-1}(x)-Fcirc f^{-1}(x)$$
                            Where $F'(x)=f(x)$. Plugging in $f(x)=tan(x)$,
                            $$I=intarctan(x)mathrm{d}x=xarctan(x)-int_0^{arctan(x)}tan(t)mathrm{d}t$$
                            Then recall that $$(-log|cos(x)|)'=tan(x)$$
                            So we have
                            $$I=xarctan(x)+log|cos(arctan(x))|$$
                            then using trig,
                            $$I=xarctan(x)-frac12log(x^2+1)$$
                            So
                            $$I_1=int_0^2arctan(x)mathrm{d}x=2arctan2-frac12log5$$
                            And
                            $$
                            begin{align}
                            I_2=&int_0^2arctan(pi x)mathrm{d}x\
                            =&frac1piint_0^{2pi}arctan(x)mathrm{d}x\
                            =&frac1pibigg(2piarctan2pi-frac12log(4pi^2+1)bigg)\
                            =&2arctan2pi-frac1{2pi}log(4pi^2+1)\
                            end{align}
                            $$
                            So
                            $$
                            begin{align}
                            int_0^2(arctanpi x-arctan x)mathrm{d}x=&I_2-I_1\
                            =&2arctan2pi-frac1{2pi}log(4pi^2+1)-2arctan2+frac12log5\
                            =&2arctan2pi-frac1{2}logsqrt[pi]{4pi^2+1}-2arctan2+frac12log5\
                            =&2(arctan2pi-arctan2)+frac12logfrac{5}{sqrt[pi]{4pi^2+1}}\
                            end{align}
                            $$






                            share|cite|improve this answer









                            $endgroup$



                            For the sake of an alternative approach, recall the formula for the integral of an inverse function
                            $$int f^{-1}(x)mathrm{d}x=xf^{-1}(x)-Fcirc f^{-1}(x)$$
                            Where $F'(x)=f(x)$. Plugging in $f(x)=tan(x)$,
                            $$I=intarctan(x)mathrm{d}x=xarctan(x)-int_0^{arctan(x)}tan(t)mathrm{d}t$$
                            Then recall that $$(-log|cos(x)|)'=tan(x)$$
                            So we have
                            $$I=xarctan(x)+log|cos(arctan(x))|$$
                            then using trig,
                            $$I=xarctan(x)-frac12log(x^2+1)$$
                            So
                            $$I_1=int_0^2arctan(x)mathrm{d}x=2arctan2-frac12log5$$
                            And
                            $$
                            begin{align}
                            I_2=&int_0^2arctan(pi x)mathrm{d}x\
                            =&frac1piint_0^{2pi}arctan(x)mathrm{d}x\
                            =&frac1pibigg(2piarctan2pi-frac12log(4pi^2+1)bigg)\
                            =&2arctan2pi-frac1{2pi}log(4pi^2+1)\
                            end{align}
                            $$
                            So
                            $$
                            begin{align}
                            int_0^2(arctanpi x-arctan x)mathrm{d}x=&I_2-I_1\
                            =&2arctan2pi-frac1{2pi}log(4pi^2+1)-2arctan2+frac12log5\
                            =&2arctan2pi-frac1{2}logsqrt[pi]{4pi^2+1}-2arctan2+frac12log5\
                            =&2(arctan2pi-arctan2)+frac12logfrac{5}{sqrt[pi]{4pi^2+1}}\
                            end{align}
                            $$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 26 '18 at 0:02









                            clathratusclathratus

                            4,419336




                            4,419336






























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