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Evaluate \int _{0}^{{2\pi }}{\frac  {d\theta }{a^{2}\sin ^{2}\theta +b^{2}\cos ^{2}\theta }}\,

Since [0,2\pi ] is a complete period of the integrand, we may take it as [\pi /2,5\pi /2] and then replace \theta by \theta +2\pi . This shows that there is no loss of generality in assuming b>a. Use double angle formulas to see that

{\frac  {1}{a^{2}\sin ^{2}\theta +b^{2}\cos ^{2}\theta }}={\frac  {2}{a^{2}(1-\cos 2\theta )+b^{2}(1+\cos 2\theta )}}={\frac  {2}{a^{2}+b^{2}+(b^{2}-a^{2})\cos 2\theta }}\,

Let 2\theta =\phi \,.

\int _{0}^{{2\pi }}{\frac  {d\theta }{a^{2}\sin ^{2}\theta +b^{2}\cos ^{2}\theta }}=\int _{0}^{{4\pi }}{\frac  {d\phi }{a^{2}+b^{2}+(b^{2}-a^{2})\cos \phi }}=2\int _{0}^{{2\pi }}{\frac  {d\phi }{a^{2}+b^{2}+(b^{2}-a^{2})\cos \phi }}\,

={\frac  {4}{i}}\int _{{|z|=1}}{\frac  {dz}{2(a^{2}+b^{2})z+(b^{2}-a^{2})(1+z^{2})}}={\frac  {4}{i}}\int _{{|z|=1}}{\frac  {dz}{(z-z_{1})(z-z_{2})(b^{2}-a^{2})}}\,

z_{1},z_{2}={\frac  {-(a^{2}+b^{2})\pm {\sqrt  {(a^{2}+b^{2})^{2}-(b^{2}-a^{2})^{2}}}}{b^{2}-a^{2}}}={\frac  {(a\mp b)^{2}}{a^{2}-b^{2}}}\,

={\frac  {a-b}{a+b}},{\frac  {a+b}{a-b}}\,

It is true that |z_{1}|<1,|z_{2}|>1\,.

The residue at z_{1}\, is

{\frac  {4}{i(b^{2}-a^{2})(z_{1}-z_{2})}}={\frac  {4}{i(b^{2}-a^{2}){\frac  {-4ab}{a^{2}-b^{2}}}}}={\frac  {1}{abi}}\,

The solution is 2\pi i{\frac  {1}{abi}}={\frac  {2\pi }{ab}}\,

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