CVRC1

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Evaluate I=\int _{0}^{{2\pi }}{\frac  {\sin ^{2}\theta }{5+4\cos \theta }}\,d\theta \,

Make the change of variables so the integral will be around the unit circle.

{\frac  {1}{z}}=e^{{-i\theta }},\cos \theta ={\frac  {e^{{i\theta }}+e^{{-i\theta }}}{2}},\sin \theta ={\frac  {e^{{i\theta }}-e^{{-i\theta }}}{2i}}\,

dz=ie^{{i\theta }},d\theta ={\frac  {dz}{iz}}\,

\cos \theta ={\frac  {1}{2}}\left(z+{\frac  {1}{z}}\right),\sin \theta ={\frac  {1}{2i}}\left(z-{\frac  {1}{z}}\right)\,

I=\int _{{|C|=1}}{\frac  {\left[{\frac  {1}{2i}}\left(z-{\frac  {1}{z}}\right)\right]^{2}}{5+4\left[{\frac  {1}{2}}\left(z+{\frac  {1}{z}}\right)\right]}}{\frac  {dz}{iz}}\,

={\frac  {-1}{4i}}\int _{C}{\frac  {z^{4}-2z^{2}+1}{z^{2}(2z^{2}+5z+2)}}dz\,

This integrand has poles z=0^{2},-2,{\frac  {-1}{2}}\,, but only 0\, and {\frac  {-1}{2}}\, are inside the unit circle.

Therefore,

I={\frac  {-1}{4i}}2\pi i\left({\mathrm  {Res}}_{{z=0}}+{\mathrm  {Res}}_{{z={\frac  {-1}{2}}}}\right)\,

{\mathrm  {Res}}_{{z=0}}=\lim _{{z\to 0}}{\frac  {d}{dz}}{\frac  {z^{4}-2z^{2}+1}{2z^{2}+5z+2}}=...={\frac  {-5}{4}}\,

{\mathrm  {Res}}_{{z={\frac  {-1}{2}}}}={\frac  {z^{4}-2z^{2}+1}{8z^{3}+15z^{2}+4z}}{\Big |}_{{z={\frac  {-1}{2}}}}={\frac  {3}{4}}\,

Finally,

I={\frac  {-1}{4i}}2\pi i\left({\frac  {-5}{4}}+{\frac  {3}{4}}\right)={\frac  {\pi }{4}}\,


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