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







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












$begingroup$


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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 5:26









Saad

19.7k92352




19.7k92352










asked Nov 4 '14 at 6:00









DiyaDiya

1,2862925




1,2862925








  • 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
3






active

oldest

votes


















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












    $begingroup$

    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









      $endgroup$













        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "69"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1005422%2fevaluating-int-02-tan-1-pi-x-tan-1-x-mathrmdx%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        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














            share|cite|improve this answer



            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






























                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Mathematics Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid …



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1005422%2fevaluating-int-02-tan-1-pi-x-tan-1-x-mathrmdx%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                -

                                Popular posts from this blog

                                Bressuire

                                Image
                                .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                                Read more

                                Cabo Verde

                                Image
                                  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
                                Read more

                                Gyllenstierna

                                Image
                                .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                                Read more