Proving $int ^{2pi} _{0} (cos(t)^{2n})={2n choose {n}}frac{2pi}{2^{2n}}$












0














using the following result



$int _{gamma}(z+ frac{1}{z})^{2n}frac{dz}{z}= {2n choose {n}}2pi i $



Prove



$int ^{2pi} _{0} (cos(t)^{2n})={2n choose {n}}frac{2pi}{2^{2n}}$



I cant see how the second part can be put in this form is there any suggestions? Also used Cauchy integral formula and binomial expansion to prove first part.






trigonometry binomial-theorem cauchy-integral-formula







share|cite|improve this question





























    0














    using the following result



    $int _{gamma}(z+ frac{1}{z})^{2n}frac{dz}{z}= {2n choose {n}}2pi i $



    Prove



    $int ^{2pi} _{0} (cos(t)^{2n})={2n choose {n}}frac{2pi}{2^{2n}}$



    I cant see how the second part can be put in this form is there any suggestions? Also used Cauchy integral formula and binomial expansion to prove first part.






    trigonometry binomial-theorem cauchy-integral-formula







    share|cite|improve this question



























      0












      0








      0







      using the following result



      $int _{gamma}(z+ frac{1}{z})^{2n}frac{dz}{z}= {2n choose {n}}2pi i $



      Prove



      $int ^{2pi} _{0} (cos(t)^{2n})={2n choose {n}}frac{2pi}{2^{2n}}$



      I cant see how the second part can be put in this form is there any suggestions? Also used Cauchy integral formula and binomial expansion to prove first part.






      trigonometry binomial-theorem cauchy-integral-formula







      share|cite|improve this question















      using the following result



      $int _{gamma}(z+ frac{1}{z})^{2n}frac{dz}{z}= {2n choose {n}}2pi i $



      Prove



      $int ^{2pi} _{0} (cos(t)^{2n})={2n choose {n}}frac{2pi}{2^{2n}}$



      I cant see how the second part can be put in this form is there any suggestions? Also used Cauchy integral formula and binomial expansion to prove first part.






      trigonometry binomial-theorem cauchy-integral-formula




      trigonometry binomial-theorem cauchy-integral-formula






      share|cite|improve this question















      share|cite|improve this question













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      edited Nov 24 at 15:25









      Bernard

      117k637111




      117k637111










      asked Nov 24 at 13:17









      frankfields

      165




      165






















          1 Answer
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          Take $gamma={(x,y)|x^2+y^2=1}.$ Then $int_gamma (z+1/z)^{2n}(1/z)dz=int_0^{2pi}(2cos t)^{2n}(cos t-isin t)(-sin t+icos t)dt=iint_0^{2pi}(2cos t)^{2n}dt={2nchoose n}2pi i.$



          And the rest follows.






          share|cite|improve this answer





















          • It's line integral, we are integrating over a curve where we have a parametric form of the curve $gamma={cos t+isin t : 0leq t<2pi}.$
            – John_Wick
            Nov 24 at 14:11












          • @John_Wick, what's the meaning of $gamma={(x,y)|(x^2+y^2=1)}$
            – Dhamnekar Winod
            Nov 24 at 14:13










          • I meant unit circle on complex plane centered at zero.
            – John_Wick
            Nov 24 at 14:16










          • services.math.duke.edu/education/ccp/materials/engin/cint1/… check here how to calculate line integral on the complex plane.
            – John_Wick
            Nov 24 at 14:47











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          Take $gamma={(x,y)|x^2+y^2=1}.$ Then $int_gamma (z+1/z)^{2n}(1/z)dz=int_0^{2pi}(2cos t)^{2n}(cos t-isin t)(-sin t+icos t)dt=iint_0^{2pi}(2cos t)^{2n}dt={2nchoose n}2pi i.$



          And the rest follows.






          share|cite|improve this answer





















          • It's line integral, we are integrating over a curve where we have a parametric form of the curve $gamma={cos t+isin t : 0leq t<2pi}.$
            – John_Wick
            Nov 24 at 14:11












          • @John_Wick, what's the meaning of $gamma={(x,y)|(x^2+y^2=1)}$
            – Dhamnekar Winod
            Nov 24 at 14:13










          • I meant unit circle on complex plane centered at zero.
            – John_Wick
            Nov 24 at 14:16










          • services.math.duke.edu/education/ccp/materials/engin/cint1/… check here how to calculate line integral on the complex plane.
            – John_Wick
            Nov 24 at 14:47
















          1














          Take $gamma={(x,y)|x^2+y^2=1}.$ Then $int_gamma (z+1/z)^{2n}(1/z)dz=int_0^{2pi}(2cos t)^{2n}(cos t-isin t)(-sin t+icos t)dt=iint_0^{2pi}(2cos t)^{2n}dt={2nchoose n}2pi i.$



          And the rest follows.






          share|cite|improve this answer





















          • It's line integral, we are integrating over a curve where we have a parametric form of the curve $gamma={cos t+isin t : 0leq t<2pi}.$
            – John_Wick
            Nov 24 at 14:11












          • @John_Wick, what's the meaning of $gamma={(x,y)|(x^2+y^2=1)}$
            – Dhamnekar Winod
            Nov 24 at 14:13










          • I meant unit circle on complex plane centered at zero.
            – John_Wick
            Nov 24 at 14:16










          • services.math.duke.edu/education/ccp/materials/engin/cint1/… check here how to calculate line integral on the complex plane.
            – John_Wick
            Nov 24 at 14:47














          1












          1








          1






          Take $gamma={(x,y)|x^2+y^2=1}.$ Then $int_gamma (z+1/z)^{2n}(1/z)dz=int_0^{2pi}(2cos t)^{2n}(cos t-isin t)(-sin t+icos t)dt=iint_0^{2pi}(2cos t)^{2n}dt={2nchoose n}2pi i.$



          And the rest follows.






          share|cite|improve this answer












          Take $gamma={(x,y)|x^2+y^2=1}.$ Then $int_gamma (z+1/z)^{2n}(1/z)dz=int_0^{2pi}(2cos t)^{2n}(cos t-isin t)(-sin t+icos t)dt=iint_0^{2pi}(2cos t)^{2n}dt={2nchoose n}2pi i.$



          And the rest follows.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 24 at 13:26









          John_Wick

          1,314111




          1,314111












          • It's line integral, we are integrating over a curve where we have a parametric form of the curve $gamma={cos t+isin t : 0leq t<2pi}.$
            – John_Wick
            Nov 24 at 14:11












          • @John_Wick, what's the meaning of $gamma={(x,y)|(x^2+y^2=1)}$
            – Dhamnekar Winod
            Nov 24 at 14:13










          • I meant unit circle on complex plane centered at zero.
            – John_Wick
            Nov 24 at 14:16










          • services.math.duke.edu/education/ccp/materials/engin/cint1/… check here how to calculate line integral on the complex plane.
            – John_Wick
            Nov 24 at 14:47


















          • It's line integral, we are integrating over a curve where we have a parametric form of the curve $gamma={cos t+isin t : 0leq t<2pi}.$
            – John_Wick
            Nov 24 at 14:11












          • @John_Wick, what's the meaning of $gamma={(x,y)|(x^2+y^2=1)}$
            – Dhamnekar Winod
            Nov 24 at 14:13










          • I meant unit circle on complex plane centered at zero.
            – John_Wick
            Nov 24 at 14:16










          • services.math.duke.edu/education/ccp/materials/engin/cint1/… check here how to calculate line integral on the complex plane.
            – John_Wick
            Nov 24 at 14:47
















          It's line integral, we are integrating over a curve where we have a parametric form of the curve $gamma={cos t+isin t : 0leq t<2pi}.$
          – John_Wick
          Nov 24 at 14:11






          It's line integral, we are integrating over a curve where we have a parametric form of the curve $gamma={cos t+isin t : 0leq t<2pi}.$
          – John_Wick
          Nov 24 at 14:11














          @John_Wick, what's the meaning of $gamma={(x,y)|(x^2+y^2=1)}$
          – Dhamnekar Winod
          Nov 24 at 14:13




          @John_Wick, what's the meaning of $gamma={(x,y)|(x^2+y^2=1)}$
          – Dhamnekar Winod
          Nov 24 at 14:13












          I meant unit circle on complex plane centered at zero.
          – John_Wick
          Nov 24 at 14:16




          I meant unit circle on complex plane centered at zero.
          – John_Wick
          Nov 24 at 14:16












          services.math.duke.edu/education/ccp/materials/engin/cint1/… check here how to calculate line integral on the complex plane.
          – John_Wick
          Nov 24 at 14:47




          services.math.duke.edu/education/ccp/materials/engin/cint1/… check here how to calculate line integral on the complex plane.
          – John_Wick
          Nov 24 at 14:47


















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