Prove that $int_0^{frac{npi}{4}} frac{1}{3sin^4(x) + 3cos^4(x) -1} dx = frac{npi}{4} $












4












$begingroup$



Prove that $$int_0^frac{npi}{4} left(frac{1}{3sin^4(x) + 3cos^4(x) -1}right) dx = frac{n pi}{4} $$
is true for all integers $n$.




Here I've looked at the case where $n=1$ as $t=tan(x)$ is injective in that domain. Then I could perhaps try to use some argument about periodicity and symmetry to extend it to all the integers.



$$
= int_0^frac{pi}{4} frac{sec^4(x)}{3tan^4(x)+3-sec^4(x)} text{d}x
= frac12 int_0^1 frac{t^2+1}{t^4-t^2+1} text{d}t
$$



Which looks a little tough but reminiscent of the well-known integral:



$$ int_0^infty frac{1}{x^4 + 2x^2cos(2a) + 1} text{d}x = frac12int_0^infty frac{x^2+1}{x^4 + 2x^2cos(2a) + 1} text{d}x = frac{pi}{4cos(a)} $$



Where we choose $a=fracpi3$. But I can't seem to adjust the bounds to get an answer. How would you evaluate this integral?










share|cite|improve this question











$endgroup$

















    4












    $begingroup$



    Prove that $$int_0^frac{npi}{4} left(frac{1}{3sin^4(x) + 3cos^4(x) -1}right) dx = frac{n pi}{4} $$
    is true for all integers $n$.




    Here I've looked at the case where $n=1$ as $t=tan(x)$ is injective in that domain. Then I could perhaps try to use some argument about periodicity and symmetry to extend it to all the integers.



    $$
    = int_0^frac{pi}{4} frac{sec^4(x)}{3tan^4(x)+3-sec^4(x)} text{d}x
    = frac12 int_0^1 frac{t^2+1}{t^4-t^2+1} text{d}t
    $$



    Which looks a little tough but reminiscent of the well-known integral:



    $$ int_0^infty frac{1}{x^4 + 2x^2cos(2a) + 1} text{d}x = frac12int_0^infty frac{x^2+1}{x^4 + 2x^2cos(2a) + 1} text{d}x = frac{pi}{4cos(a)} $$



    Where we choose $a=fracpi3$. But I can't seem to adjust the bounds to get an answer. How would you evaluate this integral?










    share|cite|improve this question











    $endgroup$















      4












      4








      4


      1



      $begingroup$



      Prove that $$int_0^frac{npi}{4} left(frac{1}{3sin^4(x) + 3cos^4(x) -1}right) dx = frac{n pi}{4} $$
      is true for all integers $n$.




      Here I've looked at the case where $n=1$ as $t=tan(x)$ is injective in that domain. Then I could perhaps try to use some argument about periodicity and symmetry to extend it to all the integers.



      $$
      = int_0^frac{pi}{4} frac{sec^4(x)}{3tan^4(x)+3-sec^4(x)} text{d}x
      = frac12 int_0^1 frac{t^2+1}{t^4-t^2+1} text{d}t
      $$



      Which looks a little tough but reminiscent of the well-known integral:



      $$ int_0^infty frac{1}{x^4 + 2x^2cos(2a) + 1} text{d}x = frac12int_0^infty frac{x^2+1}{x^4 + 2x^2cos(2a) + 1} text{d}x = frac{pi}{4cos(a)} $$



      Where we choose $a=fracpi3$. But I can't seem to adjust the bounds to get an answer. How would you evaluate this integral?










      share|cite|improve this question











      $endgroup$





      Prove that $$int_0^frac{npi}{4} left(frac{1}{3sin^4(x) + 3cos^4(x) -1}right) dx = frac{n pi}{4} $$
      is true for all integers $n$.




      Here I've looked at the case where $n=1$ as $t=tan(x)$ is injective in that domain. Then I could perhaps try to use some argument about periodicity and symmetry to extend it to all the integers.



      $$
      = int_0^frac{pi}{4} frac{sec^4(x)}{3tan^4(x)+3-sec^4(x)} text{d}x
      = frac12 int_0^1 frac{t^2+1}{t^4-t^2+1} text{d}t
      $$



      Which looks a little tough but reminiscent of the well-known integral:



      $$ int_0^infty frac{1}{x^4 + 2x^2cos(2a) + 1} text{d}x = frac12int_0^infty frac{x^2+1}{x^4 + 2x^2cos(2a) + 1} text{d}x = frac{pi}{4cos(a)} $$



      Where we choose $a=fracpi3$. But I can't seem to adjust the bounds to get an answer. How would you evaluate this integral?







      calculus integration definite-integrals trigonometric-integrals






      share|cite|improve this question















      share|cite|improve this question













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      edited Dec 2 '18 at 10:01









      Chinnapparaj R

      5,3641828




      5,3641828










      asked Dec 2 '18 at 9:11









      MintMint

      5301417




      5301417






















          4 Answers
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          3












          $begingroup$

          For $x in (0,pi/4)$,
          the integral $$frac12 int_0^1 frac{t^2+1}{t^4-t^2+1} text{d}t$$ can also be written as $$frac12 int_0^1 frac{1+1/t^2}{(t-1/t)^2+1} text{d}t$$
          Substituting $u=t-frac1t$, the integral becomes $$frac12 int_{-infty}^0 frac{1}{u^2+1} text{d}u$$ which is equal to $$frac12[arctan(u)]_{-infty}^0=pi/4$$
          Similarly for $x in (pi/4,pi/2)$, the integral becomes$$frac12 int_1^infty frac{1+1/t^2}{(t-1/t)^2+1} text{d}t$$ which is equal to$$frac12 int_0^{infty} frac{1}{v^2+1} text{d}v$$ or$$frac12[arctan(u)]_0^{infty}=pi/4$$
          And since the integrand has a period of $pi/2$ because $$sin^4x+cos^4x=1-frac12sin^22x$$
          one can easily contemplate the proof by adding the areas of the curve on and on.






          share|cite|improve this answer











          $endgroup$





















            4












            $begingroup$

            $$int_0^{pi/4}frac{1}{3sin^4(x) + 3cos^4(x) -1} dx = frac{pi}{4}$$
            can be shown by using partial fraction in OP's substituted integral.
            One can similarly show $$int_{pi/4}^{pi/2}frac{1}{3sin^4(x) + 3cos^4(x) -1} dx = frac{pi}{4}$$
            Observe that the integrand has period $fracpi2$, the problem can be easily done by using induction.






            share|cite|improve this answer











            $endgroup$





















              2












              $begingroup$

              $$3(sin^4x+cos^4x)-1=3((sin^2x+cos^2x)^2-2sin^2xcos^2x)-1=2-frac32sin^22x=frac{5+3cos4x}4$$



              and your integral is also



              $$int_0^{npi}frac{dx}{5+3cos x}=nint_0^{pi}frac{dx}{5+3cos x}$$ as the cosine is an even function.



              Using the Weierstrass substitution, the last integral is shown to be $dfracpi4$.



              $$int_0^pifrac{dx}{5+3cos4x}=int_0^inftyfrac{2,dt}{(1+t^2)left(5+3dfrac{1-t^2}{1+t^2}right)}=int_0^inftyfrac{dt}{4+t^2}.$$






              share|cite|improve this answer











              $endgroup$





















                2












                $begingroup$

                Note that the integrand function $f(x)=frac{1}{3sin^4(x) + 3cos^4(x) -1}$ is even with period $pi/2$, so
                $$int_0^frac{npi}{4} f(x) dx=
                frac{1}{2}int_{-frac{npi}{4}}^frac{npi}{4} f(x) dx=
                frac{1}{2}int_{0}^frac{npi}{2} f(x) dx
                =frac{n}{2}int_{0}^frac{pi}{2} f(x) dx.$$

                Now, following your approach, we have that
                $$frac{t^2+1}{t^4-t^2+1} =
                frac{2}{1+(2t+sqrt{3})^2} +frac{2}{1+(2t-sqrt{3})^2}.$$

                Therefore
                $$intfrac{t^2+1}{t^4-t^2+1},dt=arctan(2t+sqrt{3})+arctan(2t-sqrt{3})+c$$
                Hence, for $t=tan(x)$,
                $$int_0^frac{pi}{2} f(x) dx
                = frac12 int_0^{infty} frac{t^2+1}{t^4-t^2+1} dt=frac{1}{2}left[arctan(2t+sqrt{3})+arctan(2t-sqrt{3})right]_0^{+infty}=frac{pi}{2}.$$






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  Oh wow thanks I wouldn't have thought of partial fraction decomposition.
                  $endgroup$
                  – Mint
                  Dec 2 '18 at 9:39






                • 1




                  $begingroup$
                  @jiaminglimjm I tried to simplify the idea of your approach.
                  $endgroup$
                  – Robert Z
                  Dec 2 '18 at 10:12













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                4 Answers
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                4 Answers
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                3












                $begingroup$

                For $x in (0,pi/4)$,
                the integral $$frac12 int_0^1 frac{t^2+1}{t^4-t^2+1} text{d}t$$ can also be written as $$frac12 int_0^1 frac{1+1/t^2}{(t-1/t)^2+1} text{d}t$$
                Substituting $u=t-frac1t$, the integral becomes $$frac12 int_{-infty}^0 frac{1}{u^2+1} text{d}u$$ which is equal to $$frac12[arctan(u)]_{-infty}^0=pi/4$$
                Similarly for $x in (pi/4,pi/2)$, the integral becomes$$frac12 int_1^infty frac{1+1/t^2}{(t-1/t)^2+1} text{d}t$$ which is equal to$$frac12 int_0^{infty} frac{1}{v^2+1} text{d}v$$ or$$frac12[arctan(u)]_0^{infty}=pi/4$$
                And since the integrand has a period of $pi/2$ because $$sin^4x+cos^4x=1-frac12sin^22x$$
                one can easily contemplate the proof by adding the areas of the curve on and on.






                share|cite|improve this answer











                $endgroup$


















                  3












                  $begingroup$

                  For $x in (0,pi/4)$,
                  the integral $$frac12 int_0^1 frac{t^2+1}{t^4-t^2+1} text{d}t$$ can also be written as $$frac12 int_0^1 frac{1+1/t^2}{(t-1/t)^2+1} text{d}t$$
                  Substituting $u=t-frac1t$, the integral becomes $$frac12 int_{-infty}^0 frac{1}{u^2+1} text{d}u$$ which is equal to $$frac12[arctan(u)]_{-infty}^0=pi/4$$
                  Similarly for $x in (pi/4,pi/2)$, the integral becomes$$frac12 int_1^infty frac{1+1/t^2}{(t-1/t)^2+1} text{d}t$$ which is equal to$$frac12 int_0^{infty} frac{1}{v^2+1} text{d}v$$ or$$frac12[arctan(u)]_0^{infty}=pi/4$$
                  And since the integrand has a period of $pi/2$ because $$sin^4x+cos^4x=1-frac12sin^22x$$
                  one can easily contemplate the proof by adding the areas of the curve on and on.






                  share|cite|improve this answer











                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    For $x in (0,pi/4)$,
                    the integral $$frac12 int_0^1 frac{t^2+1}{t^4-t^2+1} text{d}t$$ can also be written as $$frac12 int_0^1 frac{1+1/t^2}{(t-1/t)^2+1} text{d}t$$
                    Substituting $u=t-frac1t$, the integral becomes $$frac12 int_{-infty}^0 frac{1}{u^2+1} text{d}u$$ which is equal to $$frac12[arctan(u)]_{-infty}^0=pi/4$$
                    Similarly for $x in (pi/4,pi/2)$, the integral becomes$$frac12 int_1^infty frac{1+1/t^2}{(t-1/t)^2+1} text{d}t$$ which is equal to$$frac12 int_0^{infty} frac{1}{v^2+1} text{d}v$$ or$$frac12[arctan(u)]_0^{infty}=pi/4$$
                    And since the integrand has a period of $pi/2$ because $$sin^4x+cos^4x=1-frac12sin^22x$$
                    one can easily contemplate the proof by adding the areas of the curve on and on.






                    share|cite|improve this answer











                    $endgroup$



                    For $x in (0,pi/4)$,
                    the integral $$frac12 int_0^1 frac{t^2+1}{t^4-t^2+1} text{d}t$$ can also be written as $$frac12 int_0^1 frac{1+1/t^2}{(t-1/t)^2+1} text{d}t$$
                    Substituting $u=t-frac1t$, the integral becomes $$frac12 int_{-infty}^0 frac{1}{u^2+1} text{d}u$$ which is equal to $$frac12[arctan(u)]_{-infty}^0=pi/4$$
                    Similarly for $x in (pi/4,pi/2)$, the integral becomes$$frac12 int_1^infty frac{1+1/t^2}{(t-1/t)^2+1} text{d}t$$ which is equal to$$frac12 int_0^{infty} frac{1}{v^2+1} text{d}v$$ or$$frac12[arctan(u)]_0^{infty}=pi/4$$
                    And since the integrand has a period of $pi/2$ because $$sin^4x+cos^4x=1-frac12sin^22x$$
                    one can easily contemplate the proof by adding the areas of the curve on and on.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 2 '18 at 10:26









                    Mint

                    5301417




                    5301417










                    answered Dec 2 '18 at 10:12









                    Sameer BahetiSameer Baheti

                    5168




                    5168























                        4












                        $begingroup$

                        $$int_0^{pi/4}frac{1}{3sin^4(x) + 3cos^4(x) -1} dx = frac{pi}{4}$$
                        can be shown by using partial fraction in OP's substituted integral.
                        One can similarly show $$int_{pi/4}^{pi/2}frac{1}{3sin^4(x) + 3cos^4(x) -1} dx = frac{pi}{4}$$
                        Observe that the integrand has period $fracpi2$, the problem can be easily done by using induction.






                        share|cite|improve this answer











                        $endgroup$


















                          4












                          $begingroup$

                          $$int_0^{pi/4}frac{1}{3sin^4(x) + 3cos^4(x) -1} dx = frac{pi}{4}$$
                          can be shown by using partial fraction in OP's substituted integral.
                          One can similarly show $$int_{pi/4}^{pi/2}frac{1}{3sin^4(x) + 3cos^4(x) -1} dx = frac{pi}{4}$$
                          Observe that the integrand has period $fracpi2$, the problem can be easily done by using induction.






                          share|cite|improve this answer











                          $endgroup$
















                            4












                            4








                            4





                            $begingroup$

                            $$int_0^{pi/4}frac{1}{3sin^4(x) + 3cos^4(x) -1} dx = frac{pi}{4}$$
                            can be shown by using partial fraction in OP's substituted integral.
                            One can similarly show $$int_{pi/4}^{pi/2}frac{1}{3sin^4(x) + 3cos^4(x) -1} dx = frac{pi}{4}$$
                            Observe that the integrand has period $fracpi2$, the problem can be easily done by using induction.






                            share|cite|improve this answer











                            $endgroup$



                            $$int_0^{pi/4}frac{1}{3sin^4(x) + 3cos^4(x) -1} dx = frac{pi}{4}$$
                            can be shown by using partial fraction in OP's substituted integral.
                            One can similarly show $$int_{pi/4}^{pi/2}frac{1}{3sin^4(x) + 3cos^4(x) -1} dx = frac{pi}{4}$$
                            Observe that the integrand has period $fracpi2$, the problem can be easily done by using induction.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 2 '18 at 9:42

























                            answered Dec 2 '18 at 9:26









                            Kemono ChenKemono Chen

                            3,0421743




                            3,0421743























                                2












                                $begingroup$

                                $$3(sin^4x+cos^4x)-1=3((sin^2x+cos^2x)^2-2sin^2xcos^2x)-1=2-frac32sin^22x=frac{5+3cos4x}4$$



                                and your integral is also



                                $$int_0^{npi}frac{dx}{5+3cos x}=nint_0^{pi}frac{dx}{5+3cos x}$$ as the cosine is an even function.



                                Using the Weierstrass substitution, the last integral is shown to be $dfracpi4$.



                                $$int_0^pifrac{dx}{5+3cos4x}=int_0^inftyfrac{2,dt}{(1+t^2)left(5+3dfrac{1-t^2}{1+t^2}right)}=int_0^inftyfrac{dt}{4+t^2}.$$






                                share|cite|improve this answer











                                $endgroup$


















                                  2












                                  $begingroup$

                                  $$3(sin^4x+cos^4x)-1=3((sin^2x+cos^2x)^2-2sin^2xcos^2x)-1=2-frac32sin^22x=frac{5+3cos4x}4$$



                                  and your integral is also



                                  $$int_0^{npi}frac{dx}{5+3cos x}=nint_0^{pi}frac{dx}{5+3cos x}$$ as the cosine is an even function.



                                  Using the Weierstrass substitution, the last integral is shown to be $dfracpi4$.



                                  $$int_0^pifrac{dx}{5+3cos4x}=int_0^inftyfrac{2,dt}{(1+t^2)left(5+3dfrac{1-t^2}{1+t^2}right)}=int_0^inftyfrac{dt}{4+t^2}.$$






                                  share|cite|improve this answer











                                  $endgroup$
















                                    2












                                    2








                                    2





                                    $begingroup$

                                    $$3(sin^4x+cos^4x)-1=3((sin^2x+cos^2x)^2-2sin^2xcos^2x)-1=2-frac32sin^22x=frac{5+3cos4x}4$$



                                    and your integral is also



                                    $$int_0^{npi}frac{dx}{5+3cos x}=nint_0^{pi}frac{dx}{5+3cos x}$$ as the cosine is an even function.



                                    Using the Weierstrass substitution, the last integral is shown to be $dfracpi4$.



                                    $$int_0^pifrac{dx}{5+3cos4x}=int_0^inftyfrac{2,dt}{(1+t^2)left(5+3dfrac{1-t^2}{1+t^2}right)}=int_0^inftyfrac{dt}{4+t^2}.$$






                                    share|cite|improve this answer











                                    $endgroup$



                                    $$3(sin^4x+cos^4x)-1=3((sin^2x+cos^2x)^2-2sin^2xcos^2x)-1=2-frac32sin^22x=frac{5+3cos4x}4$$



                                    and your integral is also



                                    $$int_0^{npi}frac{dx}{5+3cos x}=nint_0^{pi}frac{dx}{5+3cos x}$$ as the cosine is an even function.



                                    Using the Weierstrass substitution, the last integral is shown to be $dfracpi4$.



                                    $$int_0^pifrac{dx}{5+3cos4x}=int_0^inftyfrac{2,dt}{(1+t^2)left(5+3dfrac{1-t^2}{1+t^2}right)}=int_0^inftyfrac{dt}{4+t^2}.$$







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited Dec 2 '18 at 9:43

























                                    answered Dec 2 '18 at 9:37









                                    Yves DaoustYves Daoust

                                    125k671223




                                    125k671223























                                        2












                                        $begingroup$

                                        Note that the integrand function $f(x)=frac{1}{3sin^4(x) + 3cos^4(x) -1}$ is even with period $pi/2$, so
                                        $$int_0^frac{npi}{4} f(x) dx=
                                        frac{1}{2}int_{-frac{npi}{4}}^frac{npi}{4} f(x) dx=
                                        frac{1}{2}int_{0}^frac{npi}{2} f(x) dx
                                        =frac{n}{2}int_{0}^frac{pi}{2} f(x) dx.$$

                                        Now, following your approach, we have that
                                        $$frac{t^2+1}{t^4-t^2+1} =
                                        frac{2}{1+(2t+sqrt{3})^2} +frac{2}{1+(2t-sqrt{3})^2}.$$

                                        Therefore
                                        $$intfrac{t^2+1}{t^4-t^2+1},dt=arctan(2t+sqrt{3})+arctan(2t-sqrt{3})+c$$
                                        Hence, for $t=tan(x)$,
                                        $$int_0^frac{pi}{2} f(x) dx
                                        = frac12 int_0^{infty} frac{t^2+1}{t^4-t^2+1} dt=frac{1}{2}left[arctan(2t+sqrt{3})+arctan(2t-sqrt{3})right]_0^{+infty}=frac{pi}{2}.$$






                                        share|cite|improve this answer











                                        $endgroup$













                                        • $begingroup$
                                          Oh wow thanks I wouldn't have thought of partial fraction decomposition.
                                          $endgroup$
                                          – Mint
                                          Dec 2 '18 at 9:39






                                        • 1




                                          $begingroup$
                                          @jiaminglimjm I tried to simplify the idea of your approach.
                                          $endgroup$
                                          – Robert Z
                                          Dec 2 '18 at 10:12


















                                        2












                                        $begingroup$

                                        Note that the integrand function $f(x)=frac{1}{3sin^4(x) + 3cos^4(x) -1}$ is even with period $pi/2$, so
                                        $$int_0^frac{npi}{4} f(x) dx=
                                        frac{1}{2}int_{-frac{npi}{4}}^frac{npi}{4} f(x) dx=
                                        frac{1}{2}int_{0}^frac{npi}{2} f(x) dx
                                        =frac{n}{2}int_{0}^frac{pi}{2} f(x) dx.$$

                                        Now, following your approach, we have that
                                        $$frac{t^2+1}{t^4-t^2+1} =
                                        frac{2}{1+(2t+sqrt{3})^2} +frac{2}{1+(2t-sqrt{3})^2}.$$

                                        Therefore
                                        $$intfrac{t^2+1}{t^4-t^2+1},dt=arctan(2t+sqrt{3})+arctan(2t-sqrt{3})+c$$
                                        Hence, for $t=tan(x)$,
                                        $$int_0^frac{pi}{2} f(x) dx
                                        = frac12 int_0^{infty} frac{t^2+1}{t^4-t^2+1} dt=frac{1}{2}left[arctan(2t+sqrt{3})+arctan(2t-sqrt{3})right]_0^{+infty}=frac{pi}{2}.$$






                                        share|cite|improve this answer











                                        $endgroup$













                                        • $begingroup$
                                          Oh wow thanks I wouldn't have thought of partial fraction decomposition.
                                          $endgroup$
                                          – Mint
                                          Dec 2 '18 at 9:39






                                        • 1




                                          $begingroup$
                                          @jiaminglimjm I tried to simplify the idea of your approach.
                                          $endgroup$
                                          – Robert Z
                                          Dec 2 '18 at 10:12
















                                        2












                                        2








                                        2





                                        $begingroup$

                                        Note that the integrand function $f(x)=frac{1}{3sin^4(x) + 3cos^4(x) -1}$ is even with period $pi/2$, so
                                        $$int_0^frac{npi}{4} f(x) dx=
                                        frac{1}{2}int_{-frac{npi}{4}}^frac{npi}{4} f(x) dx=
                                        frac{1}{2}int_{0}^frac{npi}{2} f(x) dx
                                        =frac{n}{2}int_{0}^frac{pi}{2} f(x) dx.$$

                                        Now, following your approach, we have that
                                        $$frac{t^2+1}{t^4-t^2+1} =
                                        frac{2}{1+(2t+sqrt{3})^2} +frac{2}{1+(2t-sqrt{3})^2}.$$

                                        Therefore
                                        $$intfrac{t^2+1}{t^4-t^2+1},dt=arctan(2t+sqrt{3})+arctan(2t-sqrt{3})+c$$
                                        Hence, for $t=tan(x)$,
                                        $$int_0^frac{pi}{2} f(x) dx
                                        = frac12 int_0^{infty} frac{t^2+1}{t^4-t^2+1} dt=frac{1}{2}left[arctan(2t+sqrt{3})+arctan(2t-sqrt{3})right]_0^{+infty}=frac{pi}{2}.$$






                                        share|cite|improve this answer











                                        $endgroup$



                                        Note that the integrand function $f(x)=frac{1}{3sin^4(x) + 3cos^4(x) -1}$ is even with period $pi/2$, so
                                        $$int_0^frac{npi}{4} f(x) dx=
                                        frac{1}{2}int_{-frac{npi}{4}}^frac{npi}{4} f(x) dx=
                                        frac{1}{2}int_{0}^frac{npi}{2} f(x) dx
                                        =frac{n}{2}int_{0}^frac{pi}{2} f(x) dx.$$

                                        Now, following your approach, we have that
                                        $$frac{t^2+1}{t^4-t^2+1} =
                                        frac{2}{1+(2t+sqrt{3})^2} +frac{2}{1+(2t-sqrt{3})^2}.$$

                                        Therefore
                                        $$intfrac{t^2+1}{t^4-t^2+1},dt=arctan(2t+sqrt{3})+arctan(2t-sqrt{3})+c$$
                                        Hence, for $t=tan(x)$,
                                        $$int_0^frac{pi}{2} f(x) dx
                                        = frac12 int_0^{infty} frac{t^2+1}{t^4-t^2+1} dt=frac{1}{2}left[arctan(2t+sqrt{3})+arctan(2t-sqrt{3})right]_0^{+infty}=frac{pi}{2}.$$







                                        share|cite|improve this answer














                                        share|cite|improve this answer



                                        share|cite|improve this answer








                                        edited Dec 2 '18 at 10:22

























                                        answered Dec 2 '18 at 9:27









                                        Robert ZRobert Z

                                        95.4k1064135




                                        95.4k1064135












                                        • $begingroup$
                                          Oh wow thanks I wouldn't have thought of partial fraction decomposition.
                                          $endgroup$
                                          – Mint
                                          Dec 2 '18 at 9:39






                                        • 1




                                          $begingroup$
                                          @jiaminglimjm I tried to simplify the idea of your approach.
                                          $endgroup$
                                          – Robert Z
                                          Dec 2 '18 at 10:12




















                                        • $begingroup$
                                          Oh wow thanks I wouldn't have thought of partial fraction decomposition.
                                          $endgroup$
                                          – Mint
                                          Dec 2 '18 at 9:39






                                        • 1




                                          $begingroup$
                                          @jiaminglimjm I tried to simplify the idea of your approach.
                                          $endgroup$
                                          – Robert Z
                                          Dec 2 '18 at 10:12


















                                        $begingroup$
                                        Oh wow thanks I wouldn't have thought of partial fraction decomposition.
                                        $endgroup$
                                        – Mint
                                        Dec 2 '18 at 9:39




                                        $begingroup$
                                        Oh wow thanks I wouldn't have thought of partial fraction decomposition.
                                        $endgroup$
                                        – Mint
                                        Dec 2 '18 at 9:39




                                        1




                                        1




                                        $begingroup$
                                        @jiaminglimjm I tried to simplify the idea of your approach.
                                        $endgroup$
                                        – Robert Z
                                        Dec 2 '18 at 10:12






                                        $begingroup$
                                        @jiaminglimjm I tried to simplify the idea of your approach.
                                        $endgroup$
                                        – Robert Z
                                        Dec 2 '18 at 10:12




















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