$int_{0}^{pi}frac{sin^2(x)}{a+cos(x)},dx$












0












$begingroup$


I have been trying to solve the integral below using contour integration.
$$int_{0}^{pi}frac{sin^2(x)}{a+cos(x)}dx, quad a>1.$$



I'd appreciate to see how you would solve it, since the fact that it goes to only $pi$ and not to $2pi$ gives me some complications when solving for the residues, after using the variable change $y=2x$.






complex-analysis contour-integration







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
    $endgroup$
    – Robert Israel
    Jan 3 at 16:33










  • $begingroup$
    Oh true, ill give that a try. Thanks
    $endgroup$
    – Sam
    Jan 3 at 16:39










  • $begingroup$
    Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 3 at 16:42
















0












$begingroup$


I have been trying to solve the integral below using contour integration.
$$int_{0}^{pi}frac{sin^2(x)}{a+cos(x)}dx, quad a>1.$$



I'd appreciate to see how you would solve it, since the fact that it goes to only $pi$ and not to $2pi$ gives me some complications when solving for the residues, after using the variable change $y=2x$.






complex-analysis contour-integration







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
    $endgroup$
    – Robert Israel
    Jan 3 at 16:33










  • $begingroup$
    Oh true, ill give that a try. Thanks
    $endgroup$
    – Sam
    Jan 3 at 16:39










  • $begingroup$
    Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 3 at 16:42














0












0








0


1



$begingroup$


I have been trying to solve the integral below using contour integration.
$$int_{0}^{pi}frac{sin^2(x)}{a+cos(x)}dx, quad a>1.$$



I'd appreciate to see how you would solve it, since the fact that it goes to only $pi$ and not to $2pi$ gives me some complications when solving for the residues, after using the variable change $y=2x$.






complex-analysis contour-integration







share|cite|improve this question











$endgroup$




I have been trying to solve the integral below using contour integration.
$$int_{0}^{pi}frac{sin^2(x)}{a+cos(x)}dx, quad a>1.$$



I'd appreciate to see how you would solve it, since the fact that it goes to only $pi$ and not to $2pi$ gives me some complications when solving for the residues, after using the variable change $y=2x$.






complex-analysis contour-integration




complex-analysis contour-integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 18:36









Bernard

122k740116




122k740116










asked Jan 3 at 16:32









SamSam

394114




394114








  • 1




    $begingroup$
    I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
    $endgroup$
    – Robert Israel
    Jan 3 at 16:33










  • $begingroup$
    Oh true, ill give that a try. Thanks
    $endgroup$
    – Sam
    Jan 3 at 16:39










  • $begingroup$
    Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 3 at 16:42














  • 1




    $begingroup$
    I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
    $endgroup$
    – Robert Israel
    Jan 3 at 16:33










  • $begingroup$
    Oh true, ill give that a try. Thanks
    $endgroup$
    – Sam
    Jan 3 at 16:39










  • $begingroup$
    Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 3 at 16:42








1




1




$begingroup$
I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
$endgroup$
– Robert Israel
Jan 3 at 16:33




$begingroup$
I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
$endgroup$
– Robert Israel
Jan 3 at 16:33












$begingroup$
Oh true, ill give that a try. Thanks
$endgroup$
– Sam
Jan 3 at 16:39




$begingroup$
Oh true, ill give that a try. Thanks
$endgroup$
– Sam
Jan 3 at 16:39












$begingroup$
Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
$endgroup$
– Dr. Sonnhard Graubner
Jan 3 at 16:42




$begingroup$
Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
$endgroup$
– Dr. Sonnhard Graubner
Jan 3 at 16:42










1 Answer
1






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$begingroup$

The chances of me saying this were high: let us avoid contour integration.

By symmetry
$$ int_{0}^{pi}frac{sin^2theta}{a+costheta},dtheta = int_{0}^{pi/2}sin^2(theta)left[frac{1}{a+costheta}+frac{1}{a-costheta}right]dtheta$$
equals
$$ 2aint_{0}^{pi/2}frac{sin^2theta}{a^2-cos^2theta},dthetastackrel{thetamapstoarctan t}{=}2aint_{0}^{+infty}frac{dt}{(1+t^2)(a^2+(a^2-1)t^2)} $$
or
$$ frac{pi}{a+sqrt{a^2-1}} $$
by partial fraction decomposition.






share|cite|improve this answer









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    $begingroup$

    The chances of me saying this were high: let us avoid contour integration.

    By symmetry
    $$ int_{0}^{pi}frac{sin^2theta}{a+costheta},dtheta = int_{0}^{pi/2}sin^2(theta)left[frac{1}{a+costheta}+frac{1}{a-costheta}right]dtheta$$
    equals
    $$ 2aint_{0}^{pi/2}frac{sin^2theta}{a^2-cos^2theta},dthetastackrel{thetamapstoarctan t}{=}2aint_{0}^{+infty}frac{dt}{(1+t^2)(a^2+(a^2-1)t^2)} $$
    or
    $$ frac{pi}{a+sqrt{a^2-1}} $$
    by partial fraction decomposition.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      The chances of me saying this were high: let us avoid contour integration.

      By symmetry
      $$ int_{0}^{pi}frac{sin^2theta}{a+costheta},dtheta = int_{0}^{pi/2}sin^2(theta)left[frac{1}{a+costheta}+frac{1}{a-costheta}right]dtheta$$
      equals
      $$ 2aint_{0}^{pi/2}frac{sin^2theta}{a^2-cos^2theta},dthetastackrel{thetamapstoarctan t}{=}2aint_{0}^{+infty}frac{dt}{(1+t^2)(a^2+(a^2-1)t^2)} $$
      or
      $$ frac{pi}{a+sqrt{a^2-1}} $$
      by partial fraction decomposition.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        The chances of me saying this were high: let us avoid contour integration.

        By symmetry
        $$ int_{0}^{pi}frac{sin^2theta}{a+costheta},dtheta = int_{0}^{pi/2}sin^2(theta)left[frac{1}{a+costheta}+frac{1}{a-costheta}right]dtheta$$
        equals
        $$ 2aint_{0}^{pi/2}frac{sin^2theta}{a^2-cos^2theta},dthetastackrel{thetamapstoarctan t}{=}2aint_{0}^{+infty}frac{dt}{(1+t^2)(a^2+(a^2-1)t^2)} $$
        or
        $$ frac{pi}{a+sqrt{a^2-1}} $$
        by partial fraction decomposition.






        share|cite|improve this answer









        $endgroup$



        The chances of me saying this were high: let us avoid contour integration.

        By symmetry
        $$ int_{0}^{pi}frac{sin^2theta}{a+costheta},dtheta = int_{0}^{pi/2}sin^2(theta)left[frac{1}{a+costheta}+frac{1}{a-costheta}right]dtheta$$
        equals
        $$ 2aint_{0}^{pi/2}frac{sin^2theta}{a^2-cos^2theta},dthetastackrel{thetamapstoarctan t}{=}2aint_{0}^{+infty}frac{dt}{(1+t^2)(a^2+(a^2-1)t^2)} $$
        or
        $$ frac{pi}{a+sqrt{a^2-1}} $$
        by partial fraction decomposition.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 3 at 16:51









        Jack D'AurizioJack D'Aurizio

        290k33283664




        290k33283664






























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