$int_{0}^{2pi} frac{cos{3theta}}{5-4cos{theta}},d{theta}$ by using real integration












2












$begingroup$


How to evaluate $int_{0}^{2pi} frac{cos{3theta}}{5-4cos{theta}},d{theta}$
by using complex integration?



I assume $z=e^{i{theta}}$, $frac{1}{iz}dz=d{theta}$,
$$cos{theta}=frac{z+z^{-1}}{2}
quadmbox{and}quad cos3{theta}=frac{(z+z^{-1})(z^2+z^{-2}-1)}{2}.$$
Hence
$$frac{1}{2i}oint_{|z|=1} frac{{(z+z^{-1})(z^2+z^{-2}-1)}}{(5-4frac{z+z^{-1}}{2})z},dz$$
and I'm stuck in this. Could you give me a hints or solution?






integration contour-integration complex-integration







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    I guess you meant “complex” in the title of the question, not “real”?
    $endgroup$
    – Hans Lundmark
    Nov 8 '16 at 12:42










  • $begingroup$
    I'm studying example of Radius integration of real integrals part in engineering mathmatics. so I put that title as a connectivity
    $endgroup$
    – matthew
    Nov 8 '16 at 12:46
















2












$begingroup$


How to evaluate $int_{0}^{2pi} frac{cos{3theta}}{5-4cos{theta}},d{theta}$
by using complex integration?



I assume $z=e^{i{theta}}$, $frac{1}{iz}dz=d{theta}$,
$$cos{theta}=frac{z+z^{-1}}{2}
quadmbox{and}quad cos3{theta}=frac{(z+z^{-1})(z^2+z^{-2}-1)}{2}.$$
Hence
$$frac{1}{2i}oint_{|z|=1} frac{{(z+z^{-1})(z^2+z^{-2}-1)}}{(5-4frac{z+z^{-1}}{2})z},dz$$
and I'm stuck in this. Could you give me a hints or solution?






integration contour-integration complex-integration







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    I guess you meant “complex” in the title of the question, not “real”?
    $endgroup$
    – Hans Lundmark
    Nov 8 '16 at 12:42










  • $begingroup$
    I'm studying example of Radius integration of real integrals part in engineering mathmatics. so I put that title as a connectivity
    $endgroup$
    – matthew
    Nov 8 '16 at 12:46














2












2








2


1



$begingroup$


How to evaluate $int_{0}^{2pi} frac{cos{3theta}}{5-4cos{theta}},d{theta}$
by using complex integration?



I assume $z=e^{i{theta}}$, $frac{1}{iz}dz=d{theta}$,
$$cos{theta}=frac{z+z^{-1}}{2}
quadmbox{and}quad cos3{theta}=frac{(z+z^{-1})(z^2+z^{-2}-1)}{2}.$$
Hence
$$frac{1}{2i}oint_{|z|=1} frac{{(z+z^{-1})(z^2+z^{-2}-1)}}{(5-4frac{z+z^{-1}}{2})z},dz$$
and I'm stuck in this. Could you give me a hints or solution?






integration contour-integration complex-integration







share|cite|improve this question











$endgroup$




How to evaluate $int_{0}^{2pi} frac{cos{3theta}}{5-4cos{theta}},d{theta}$
by using complex integration?



I assume $z=e^{i{theta}}$, $frac{1}{iz}dz=d{theta}$,
$$cos{theta}=frac{z+z^{-1}}{2}
quadmbox{and}quad cos3{theta}=frac{(z+z^{-1})(z^2+z^{-2}-1)}{2}.$$
Hence
$$frac{1}{2i}oint_{|z|=1} frac{{(z+z^{-1})(z^2+z^{-2}-1)}}{(5-4frac{z+z^{-1}}{2})z},dz$$
and I'm stuck in this. Could you give me a hints or solution?






integration contour-integration complex-integration




integration contour-integration complex-integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 8 '16 at 12:30









Robert Z

98.5k1068139




98.5k1068139










asked Nov 8 '16 at 11:58









matthewmatthew

709




709








  • 3




    $begingroup$
    I guess you meant “complex” in the title of the question, not “real”?
    $endgroup$
    – Hans Lundmark
    Nov 8 '16 at 12:42










  • $begingroup$
    I'm studying example of Radius integration of real integrals part in engineering mathmatics. so I put that title as a connectivity
    $endgroup$
    – matthew
    Nov 8 '16 at 12:46














  • 3




    $begingroup$
    I guess you meant “complex” in the title of the question, not “real”?
    $endgroup$
    – Hans Lundmark
    Nov 8 '16 at 12:42










  • $begingroup$
    I'm studying example of Radius integration of real integrals part in engineering mathmatics. so I put that title as a connectivity
    $endgroup$
    – matthew
    Nov 8 '16 at 12:46








3




3




$begingroup$
I guess you meant “complex” in the title of the question, not “real”?
$endgroup$
– Hans Lundmark
Nov 8 '16 at 12:42




$begingroup$
I guess you meant “complex” in the title of the question, not “real”?
$endgroup$
– Hans Lundmark
Nov 8 '16 at 12:42












$begingroup$
I'm studying example of Radius integration of real integrals part in engineering mathmatics. so I put that title as a connectivity
$endgroup$
– matthew
Nov 8 '16 at 12:46




$begingroup$
I'm studying example of Radius integration of real integrals part in engineering mathmatics. so I put that title as a connectivity
$endgroup$
– matthew
Nov 8 '16 at 12:46










1 Answer
1






active

oldest

votes


















2












$begingroup$

You are on the right track. Now you should use the Residue theorem and find the residues of the integrand at 0 and at $1/2$ (the poles inside the unit circle. The final result should be $frac{pi}{12}$.
$$int_{0}^{2pi} frac{cos 3theta }{5-4cos{theta}},d{theta}=
frac{i}{4}int_{|z|=1}frac{z^6+1
}{z^3(z-frac{1}{2})(z-2)},dz=-frac{pi}{2}left(mbox{Res}(f,0)+mbox{Res}(f,1/2)right)$$
where $f$ is the integrand function.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How did you get z at 0 and$frac{1}{2}$?? I mean I'm stuck in calculation to find z..
    $endgroup$
    – matthew
    Nov 8 '16 at 12:20








  • 1




    $begingroup$
    @matthew Just simplify and factor your integrand function. I edited my answer with more details.
    $endgroup$
    – Robert Z
    Nov 8 '16 at 12:25











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

You are on the right track. Now you should use the Residue theorem and find the residues of the integrand at 0 and at $1/2$ (the poles inside the unit circle. The final result should be $frac{pi}{12}$.
$$int_{0}^{2pi} frac{cos 3theta }{5-4cos{theta}},d{theta}=
frac{i}{4}int_{|z|=1}frac{z^6+1
}{z^3(z-frac{1}{2})(z-2)},dz=-frac{pi}{2}left(mbox{Res}(f,0)+mbox{Res}(f,1/2)right)$$
where $f$ is the integrand function.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How did you get z at 0 and$frac{1}{2}$?? I mean I'm stuck in calculation to find z..
    $endgroup$
    – matthew
    Nov 8 '16 at 12:20








  • 1




    $begingroup$
    @matthew Just simplify and factor your integrand function. I edited my answer with more details.
    $endgroup$
    – Robert Z
    Nov 8 '16 at 12:25
















2












$begingroup$

You are on the right track. Now you should use the Residue theorem and find the residues of the integrand at 0 and at $1/2$ (the poles inside the unit circle. The final result should be $frac{pi}{12}$.
$$int_{0}^{2pi} frac{cos 3theta }{5-4cos{theta}},d{theta}=
frac{i}{4}int_{|z|=1}frac{z^6+1
}{z^3(z-frac{1}{2})(z-2)},dz=-frac{pi}{2}left(mbox{Res}(f,0)+mbox{Res}(f,1/2)right)$$
where $f$ is the integrand function.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How did you get z at 0 and$frac{1}{2}$?? I mean I'm stuck in calculation to find z..
    $endgroup$
    – matthew
    Nov 8 '16 at 12:20








  • 1




    $begingroup$
    @matthew Just simplify and factor your integrand function. I edited my answer with more details.
    $endgroup$
    – Robert Z
    Nov 8 '16 at 12:25














2












2








2





$begingroup$

You are on the right track. Now you should use the Residue theorem and find the residues of the integrand at 0 and at $1/2$ (the poles inside the unit circle. The final result should be $frac{pi}{12}$.
$$int_{0}^{2pi} frac{cos 3theta }{5-4cos{theta}},d{theta}=
frac{i}{4}int_{|z|=1}frac{z^6+1
}{z^3(z-frac{1}{2})(z-2)},dz=-frac{pi}{2}left(mbox{Res}(f,0)+mbox{Res}(f,1/2)right)$$
where $f$ is the integrand function.






share|cite|improve this answer











$endgroup$



You are on the right track. Now you should use the Residue theorem and find the residues of the integrand at 0 and at $1/2$ (the poles inside the unit circle. The final result should be $frac{pi}{12}$.
$$int_{0}^{2pi} frac{cos 3theta }{5-4cos{theta}},d{theta}=
frac{i}{4}int_{|z|=1}frac{z^6+1
}{z^3(z-frac{1}{2})(z-2)},dz=-frac{pi}{2}left(mbox{Res}(f,0)+mbox{Res}(f,1/2)right)$$
where $f$ is the integrand function.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 8 '16 at 12:33

























answered Nov 8 '16 at 12:16









Robert ZRobert Z

98.5k1068139




98.5k1068139












  • $begingroup$
    How did you get z at 0 and$frac{1}{2}$?? I mean I'm stuck in calculation to find z..
    $endgroup$
    – matthew
    Nov 8 '16 at 12:20








  • 1




    $begingroup$
    @matthew Just simplify and factor your integrand function. I edited my answer with more details.
    $endgroup$
    – Robert Z
    Nov 8 '16 at 12:25


















  • $begingroup$
    How did you get z at 0 and$frac{1}{2}$?? I mean I'm stuck in calculation to find z..
    $endgroup$
    – matthew
    Nov 8 '16 at 12:20








  • 1




    $begingroup$
    @matthew Just simplify and factor your integrand function. I edited my answer with more details.
    $endgroup$
    – Robert Z
    Nov 8 '16 at 12:25
















$begingroup$
How did you get z at 0 and$frac{1}{2}$?? I mean I'm stuck in calculation to find z..
$endgroup$
– matthew
Nov 8 '16 at 12:20






$begingroup$
How did you get z at 0 and$frac{1}{2}$?? I mean I'm stuck in calculation to find z..
$endgroup$
– matthew
Nov 8 '16 at 12:20






1




1




$begingroup$
@matthew Just simplify and factor your integrand function. I edited my answer with more details.
$endgroup$
– Robert Z
Nov 8 '16 at 12:25




$begingroup$
@matthew Just simplify and factor your integrand function. I edited my answer with more details.
$endgroup$
– Robert Z
Nov 8 '16 at 12:25


















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