Solving integral using residues












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I'm trying to find the value of the integral $int_0^{2pi}frac{cos^2u}{2-sin u}du$ using the substitution: $cos u =frac{1}{2}(z+frac{1}{z})$ and $sin u = frac{1}{2i}(z-frac{1}{z})$. Making the substitution, we obtain $int_C(z^2+1)^2/(2z^2(z^2+4iz-1)dz$ where C denotes the unit circle. I leave out the steps inbetween because I do not believe the error was made there.

Next, I found the roots of the polynomial in the denominator, which are the poles of the integrand. I found these to be $z_0=0$, $z_1=i(-2+sqrt3)$, $z_2=i(-2-sqrt3)$. Only $z_0$ and $z_1$ lie inside the unit circle, so the value of the integral will be $2pi i[Res(0)+Res(i(-2+sqrt3))]$. $Res(0)$ is easily obtained. $Res(i(2-sqrt3))$ should be too, since $z_1$ is a first order pole, so I can just take $lim_{zto z_1}[f(z)(z-z_1)]$. The results are $$Res(0)= lim_{zto 0}frac{d}{dz}z^2f(z)=...=-2i$$ and $$Res(i(sqrt3-2))=lim_{zto z_1}(z-z_1)f(z)=...=-3/2+sqrt3$$
What mistake did I make? My answer turns out to be $2pi i(-2i-3/2+sqrt3)$, the solution is $-2pi(sqrt3-2)$.






complex-analysis residue-calculus







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I'm trying to find the value of the integral $int_0^{2pi}frac{cos^2u}{2-sin u}du$ using the substitution: $cos u =frac{1}{2}(z+frac{1}{z})$ and $sin u = frac{1}{2i}(z-frac{1}{z})$. Making the substitution, we obtain $int_C(z^2+1)^2/(2z^2(z^2+4iz-1)dz$ where C denotes the unit circle. I leave out the steps inbetween because I do not believe the error was made there.

    Next, I found the roots of the polynomial in the denominator, which are the poles of the integrand. I found these to be $z_0=0$, $z_1=i(-2+sqrt3)$, $z_2=i(-2-sqrt3)$. Only $z_0$ and $z_1$ lie inside the unit circle, so the value of the integral will be $2pi i[Res(0)+Res(i(-2+sqrt3))]$. $Res(0)$ is easily obtained. $Res(i(2-sqrt3))$ should be too, since $z_1$ is a first order pole, so I can just take $lim_{zto z_1}[f(z)(z-z_1)]$. The results are $$Res(0)= lim_{zto 0}frac{d}{dz}z^2f(z)=...=-2i$$ and $$Res(i(sqrt3-2))=lim_{zto z_1}(z-z_1)f(z)=...=-3/2+sqrt3$$
    What mistake did I make? My answer turns out to be $2pi i(-2i-3/2+sqrt3)$, the solution is $-2pi(sqrt3-2)$.






    complex-analysis residue-calculus







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I'm trying to find the value of the integral $int_0^{2pi}frac{cos^2u}{2-sin u}du$ using the substitution: $cos u =frac{1}{2}(z+frac{1}{z})$ and $sin u = frac{1}{2i}(z-frac{1}{z})$. Making the substitution, we obtain $int_C(z^2+1)^2/(2z^2(z^2+4iz-1)dz$ where C denotes the unit circle. I leave out the steps inbetween because I do not believe the error was made there.

      Next, I found the roots of the polynomial in the denominator, which are the poles of the integrand. I found these to be $z_0=0$, $z_1=i(-2+sqrt3)$, $z_2=i(-2-sqrt3)$. Only $z_0$ and $z_1$ lie inside the unit circle, so the value of the integral will be $2pi i[Res(0)+Res(i(-2+sqrt3))]$. $Res(0)$ is easily obtained. $Res(i(2-sqrt3))$ should be too, since $z_1$ is a first order pole, so I can just take $lim_{zto z_1}[f(z)(z-z_1)]$. The results are $$Res(0)= lim_{zto 0}frac{d}{dz}z^2f(z)=...=-2i$$ and $$Res(i(sqrt3-2))=lim_{zto z_1}(z-z_1)f(z)=...=-3/2+sqrt3$$
      What mistake did I make? My answer turns out to be $2pi i(-2i-3/2+sqrt3)$, the solution is $-2pi(sqrt3-2)$.






      complex-analysis residue-calculus







      share|cite|improve this question









      $endgroup$




      I'm trying to find the value of the integral $int_0^{2pi}frac{cos^2u}{2-sin u}du$ using the substitution: $cos u =frac{1}{2}(z+frac{1}{z})$ and $sin u = frac{1}{2i}(z-frac{1}{z})$. Making the substitution, we obtain $int_C(z^2+1)^2/(2z^2(z^2+4iz-1)dz$ where C denotes the unit circle. I leave out the steps inbetween because I do not believe the error was made there.

      Next, I found the roots of the polynomial in the denominator, which are the poles of the integrand. I found these to be $z_0=0$, $z_1=i(-2+sqrt3)$, $z_2=i(-2-sqrt3)$. Only $z_0$ and $z_1$ lie inside the unit circle, so the value of the integral will be $2pi i[Res(0)+Res(i(-2+sqrt3))]$. $Res(0)$ is easily obtained. $Res(i(2-sqrt3))$ should be too, since $z_1$ is a first order pole, so I can just take $lim_{zto z_1}[f(z)(z-z_1)]$. The results are $$Res(0)= lim_{zto 0}frac{d}{dz}z^2f(z)=...=-2i$$ and $$Res(i(sqrt3-2))=lim_{zto z_1}(z-z_1)f(z)=...=-3/2+sqrt3$$
      What mistake did I make? My answer turns out to be $2pi i(-2i-3/2+sqrt3)$, the solution is $-2pi(sqrt3-2)$.






      complex-analysis residue-calculus




      complex-analysis residue-calculus






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      asked Jan 7 at 18:54









      user569579user569579

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          $begingroup$

          Let $tanfrac{u}{2}=x$.



          Thus, $$dx=frac{1}{2cos^2frac{u}{2}}du=frac{1}{2}(1+x^2)du,$$ which gives
          $$du=frac{2}{1+x^2}dx.$$
          Hence, $$intfrac{cos^2u}{2-sin{u}}du=intleft(2+sin{u}-frac{3}{2-sin{u}}right)du=$$
          $$=2u-cos{u}-3intfrac{1}{2-frac{2x}{1+x^2}}frac{2}{1+x^2}dx=$$
          $$=2u-cos{u}-3intfrac{1}{1-x+x^2}dx=2u-cos{u}-3intfrac{1}{left(x-frac{1}{2}right)^2+left(frac{sqrt3}{2}right)^2}dx=$$
          $$=2u-cos{u}-2sqrt3arctanfrac{2tanfrac{u}{2}-1}{sqrt3}+C.$$
          Can you end it now?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I probably could, but I am trying to solve this using the Residue Theorem. If I want to use that, is my approach correct? Either way, thank you for your help.
            $endgroup$
            – user569579
            Jan 7 at 21:29



















          0












          $begingroup$

          After the substitution $z = e^{ix}$, you should be getting the integral



          $$ int_{0}^{2pi}frac{cos^{2}x}{2-sin x},mathrm{d}x = int_{C}frac{(z^{2}+1)^{2}}{4z^{2}}frac{mathrm{d}z}{izleft(2-frac{1}{2i}(z-frac{1}{z})right)} = -frac{1}{2}int_{C}frac{(z^{2}+1)^{2}}{z^{2}(z^{2}-4iz-1)},mathrm{d}z.$$



          Since the original integral is clearly real, your residues need to be purely imaginary. If they are not, then something went wrong in the computation.






          share|cite|improve this answer











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            2 Answers
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            2 Answers
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            active

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            1












            $begingroup$

            Let $tanfrac{u}{2}=x$.



            Thus, $$dx=frac{1}{2cos^2frac{u}{2}}du=frac{1}{2}(1+x^2)du,$$ which gives
            $$du=frac{2}{1+x^2}dx.$$
            Hence, $$intfrac{cos^2u}{2-sin{u}}du=intleft(2+sin{u}-frac{3}{2-sin{u}}right)du=$$
            $$=2u-cos{u}-3intfrac{1}{2-frac{2x}{1+x^2}}frac{2}{1+x^2}dx=$$
            $$=2u-cos{u}-3intfrac{1}{1-x+x^2}dx=2u-cos{u}-3intfrac{1}{left(x-frac{1}{2}right)^2+left(frac{sqrt3}{2}right)^2}dx=$$
            $$=2u-cos{u}-2sqrt3arctanfrac{2tanfrac{u}{2}-1}{sqrt3}+C.$$
            Can you end it now?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I probably could, but I am trying to solve this using the Residue Theorem. If I want to use that, is my approach correct? Either way, thank you for your help.
              $endgroup$
              – user569579
              Jan 7 at 21:29
















            1












            $begingroup$

            Let $tanfrac{u}{2}=x$.



            Thus, $$dx=frac{1}{2cos^2frac{u}{2}}du=frac{1}{2}(1+x^2)du,$$ which gives
            $$du=frac{2}{1+x^2}dx.$$
            Hence, $$intfrac{cos^2u}{2-sin{u}}du=intleft(2+sin{u}-frac{3}{2-sin{u}}right)du=$$
            $$=2u-cos{u}-3intfrac{1}{2-frac{2x}{1+x^2}}frac{2}{1+x^2}dx=$$
            $$=2u-cos{u}-3intfrac{1}{1-x+x^2}dx=2u-cos{u}-3intfrac{1}{left(x-frac{1}{2}right)^2+left(frac{sqrt3}{2}right)^2}dx=$$
            $$=2u-cos{u}-2sqrt3arctanfrac{2tanfrac{u}{2}-1}{sqrt3}+C.$$
            Can you end it now?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I probably could, but I am trying to solve this using the Residue Theorem. If I want to use that, is my approach correct? Either way, thank you for your help.
              $endgroup$
              – user569579
              Jan 7 at 21:29














            1












            1








            1





            $begingroup$

            Let $tanfrac{u}{2}=x$.



            Thus, $$dx=frac{1}{2cos^2frac{u}{2}}du=frac{1}{2}(1+x^2)du,$$ which gives
            $$du=frac{2}{1+x^2}dx.$$
            Hence, $$intfrac{cos^2u}{2-sin{u}}du=intleft(2+sin{u}-frac{3}{2-sin{u}}right)du=$$
            $$=2u-cos{u}-3intfrac{1}{2-frac{2x}{1+x^2}}frac{2}{1+x^2}dx=$$
            $$=2u-cos{u}-3intfrac{1}{1-x+x^2}dx=2u-cos{u}-3intfrac{1}{left(x-frac{1}{2}right)^2+left(frac{sqrt3}{2}right)^2}dx=$$
            $$=2u-cos{u}-2sqrt3arctanfrac{2tanfrac{u}{2}-1}{sqrt3}+C.$$
            Can you end it now?






            share|cite|improve this answer









            $endgroup$



            Let $tanfrac{u}{2}=x$.



            Thus, $$dx=frac{1}{2cos^2frac{u}{2}}du=frac{1}{2}(1+x^2)du,$$ which gives
            $$du=frac{2}{1+x^2}dx.$$
            Hence, $$intfrac{cos^2u}{2-sin{u}}du=intleft(2+sin{u}-frac{3}{2-sin{u}}right)du=$$
            $$=2u-cos{u}-3intfrac{1}{2-frac{2x}{1+x^2}}frac{2}{1+x^2}dx=$$
            $$=2u-cos{u}-3intfrac{1}{1-x+x^2}dx=2u-cos{u}-3intfrac{1}{left(x-frac{1}{2}right)^2+left(frac{sqrt3}{2}right)^2}dx=$$
            $$=2u-cos{u}-2sqrt3arctanfrac{2tanfrac{u}{2}-1}{sqrt3}+C.$$
            Can you end it now?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 7 at 21:19









            Michael RozenbergMichael Rozenberg

            111k1896201




            111k1896201












            • $begingroup$
              I probably could, but I am trying to solve this using the Residue Theorem. If I want to use that, is my approach correct? Either way, thank you for your help.
              $endgroup$
              – user569579
              Jan 7 at 21:29


















            • $begingroup$
              I probably could, but I am trying to solve this using the Residue Theorem. If I want to use that, is my approach correct? Either way, thank you for your help.
              $endgroup$
              – user569579
              Jan 7 at 21:29
















            $begingroup$
            I probably could, but I am trying to solve this using the Residue Theorem. If I want to use that, is my approach correct? Either way, thank you for your help.
            $endgroup$
            – user569579
            Jan 7 at 21:29




            $begingroup$
            I probably could, but I am trying to solve this using the Residue Theorem. If I want to use that, is my approach correct? Either way, thank you for your help.
            $endgroup$
            – user569579
            Jan 7 at 21:29











            0












            $begingroup$

            After the substitution $z = e^{ix}$, you should be getting the integral



            $$ int_{0}^{2pi}frac{cos^{2}x}{2-sin x},mathrm{d}x = int_{C}frac{(z^{2}+1)^{2}}{4z^{2}}frac{mathrm{d}z}{izleft(2-frac{1}{2i}(z-frac{1}{z})right)} = -frac{1}{2}int_{C}frac{(z^{2}+1)^{2}}{z^{2}(z^{2}-4iz-1)},mathrm{d}z.$$



            Since the original integral is clearly real, your residues need to be purely imaginary. If they are not, then something went wrong in the computation.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              After the substitution $z = e^{ix}$, you should be getting the integral



              $$ int_{0}^{2pi}frac{cos^{2}x}{2-sin x},mathrm{d}x = int_{C}frac{(z^{2}+1)^{2}}{4z^{2}}frac{mathrm{d}z}{izleft(2-frac{1}{2i}(z-frac{1}{z})right)} = -frac{1}{2}int_{C}frac{(z^{2}+1)^{2}}{z^{2}(z^{2}-4iz-1)},mathrm{d}z.$$



              Since the original integral is clearly real, your residues need to be purely imaginary. If they are not, then something went wrong in the computation.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                After the substitution $z = e^{ix}$, you should be getting the integral



                $$ int_{0}^{2pi}frac{cos^{2}x}{2-sin x},mathrm{d}x = int_{C}frac{(z^{2}+1)^{2}}{4z^{2}}frac{mathrm{d}z}{izleft(2-frac{1}{2i}(z-frac{1}{z})right)} = -frac{1}{2}int_{C}frac{(z^{2}+1)^{2}}{z^{2}(z^{2}-4iz-1)},mathrm{d}z.$$



                Since the original integral is clearly real, your residues need to be purely imaginary. If they are not, then something went wrong in the computation.






                share|cite|improve this answer











                $endgroup$



                After the substitution $z = e^{ix}$, you should be getting the integral



                $$ int_{0}^{2pi}frac{cos^{2}x}{2-sin x},mathrm{d}x = int_{C}frac{(z^{2}+1)^{2}}{4z^{2}}frac{mathrm{d}z}{izleft(2-frac{1}{2i}(z-frac{1}{z})right)} = -frac{1}{2}int_{C}frac{(z^{2}+1)^{2}}{z^{2}(z^{2}-4iz-1)},mathrm{d}z.$$



                Since the original integral is clearly real, your residues need to be purely imaginary. If they are not, then something went wrong in the computation.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 9 at 13:36

























                answered Jan 9 at 5:54









                IninterrompueIninterrompue

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                69019






























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