Finding $frac{1}{2pi}int_{0}^{2pi}phi^prime(x) dx$, where $phi(x)=arctanfrac{3cos x}{4(cos x+sin x)}$. Why...












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I'm tasked with the following problem:



Evaluate
$$I_C=frac{1}{2pi}int_{0}^{2pi}left(frac{d}{dtheta}phi(theta)right) dtheta,quadtext{where}; phi(theta)=arctanleft[frac{3cos(theta)}{4(cos(theta)+sin(theta))}right]$$



Am I correct in assuming that this is simply $phi(2pi)-phi(0)$? When I do that I get zero, but when I take the derivative, then evaluate the integral I get -1. How do I use the fundamental theorem of calculus to get the correct answer?










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    1












    $begingroup$


    I'm tasked with the following problem:



    Evaluate
    $$I_C=frac{1}{2pi}int_{0}^{2pi}left(frac{d}{dtheta}phi(theta)right) dtheta,quadtext{where}; phi(theta)=arctanleft[frac{3cos(theta)}{4(cos(theta)+sin(theta))}right]$$



    Am I correct in assuming that this is simply $phi(2pi)-phi(0)$? When I do that I get zero, but when I take the derivative, then evaluate the integral I get -1. How do I use the fundamental theorem of calculus to get the correct answer?










    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I'm tasked with the following problem:



      Evaluate
      $$I_C=frac{1}{2pi}int_{0}^{2pi}left(frac{d}{dtheta}phi(theta)right) dtheta,quadtext{where}; phi(theta)=arctanleft[frac{3cos(theta)}{4(cos(theta)+sin(theta))}right]$$



      Am I correct in assuming that this is simply $phi(2pi)-phi(0)$? When I do that I get zero, but when I take the derivative, then evaluate the integral I get -1. How do I use the fundamental theorem of calculus to get the correct answer?










      share|cite|improve this question











      $endgroup$




      I'm tasked with the following problem:



      Evaluate
      $$I_C=frac{1}{2pi}int_{0}^{2pi}left(frac{d}{dtheta}phi(theta)right) dtheta,quadtext{where}; phi(theta)=arctanleft[frac{3cos(theta)}{4(cos(theta)+sin(theta))}right]$$



      Am I correct in assuming that this is simply $phi(2pi)-phi(0)$? When I do that I get zero, but when I take the derivative, then evaluate the integral I get -1. How do I use the fundamental theorem of calculus to get the correct answer?







      calculus






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      edited Dec 11 '18 at 9:10









      Blue

      48k870153




      48k870153










      asked Dec 10 '18 at 6:49









      PeetriusPeetrius

      409111




      409111






















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          $begingroup$

          Hint: what happens at $theta =3pi/4, 7pi/4$? See the graph if required.






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            $begingroup$

            Hint: what happens at $theta =3pi/4, 7pi/4$? See the graph if required.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              Hint: what happens at $theta =3pi/4, 7pi/4$? See the graph if required.






              share|cite|improve this answer











              $endgroup$
















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                $begingroup$

                Hint: what happens at $theta =3pi/4, 7pi/4$? See the graph if required.






                share|cite|improve this answer











                $endgroup$



                Hint: what happens at $theta =3pi/4, 7pi/4$? See the graph if required.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 11 '18 at 8:40

























                answered Dec 10 '18 at 8:28









                Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

                34.2k42871




                34.2k42871






























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