Question

Evaluate the integral using residue theorem ∫sin(3x)/(5-3cos(x)) where limits of integration are from zero to 2π.

Answer 1

$\int\limits_0^{2\pi } {\frac{{\sin (3x)}}{{5 - 3\cos x}}dx = \int\limits_0^{2\pi } {\frac{{3\sin x - 4{{\sin }^3}x}}{{5 - 3\cos x}}} dx = \int\limits_{|z| = 1} {R\left( {\frac{1}{2}\left( {z + \frac{1}{z}} \right),\,\,\frac{1}{{2i}}\left( {z - \frac{1}{z}} \right)} \right)} } \frac{{dz}}{{iz}} = \int\limits_{|z| = 1} {\frac{{3 \cdot \frac{1}{{2i}}\left( {z - \frac{1}{z}} \right) - 4{{\left( {\frac{1}{{2i}}\left( {z - \frac{1}{z}} \right)} \right)}^3}}}{{5 - 3 \cdot \frac{1}{2}\left( {z + \frac{1}{z}} \right)}}} \frac{{dz}}{{iz}} = \int\limits_{|z| = 1} {\frac{{\frac{3}{{2i}} \cdot \frac{{{z^2} - 1}}{z} - 4 \cdot \left( { - \frac{1}{{8i}}\left( {{z^3} - \frac{1}{{{z^3}}} - 3z + \frac{3}{z}} \right)} \right)}}{{5 - \frac{3}{2} \cdot \frac{{{z^2} + 1}}{z}}}} \frac{{dz}}{{iz}} = \int\limits_{|z| = 1} {\frac{{\frac{3}{{2i}} \cdot \frac{{{z^2} - 1}}{z} + \frac{1}{{2i}} \cdot \frac{{{z^6} - 1 - 3{z^4} + 3{z^2}}}{{{z^3}}}}}{{\frac{{10z - 3{z^2} - 3}}{{2z}}}}} \frac{{dz}}{{iz}} = \frac{1}{{2i}}\int\limits_{|z| = 1} {\frac{{\frac{{3{z^4} - 3{z^2} + {z^6} - 1 - 3{z^4} + 3{z^2}}}{{{z^3}}}}}{{\frac{{10z - 3{z^2} - 3}}{{2z}}}}} \frac{{dz}}{{iz}} = \frac{1}{i}\int\limits_{|z| = 1} {\frac{{\left( {3{z^4} - 3{z^2} + {z^6} - 1 - 3{z^4} + 3{z^2}} \right)}}{{{z^2}\left( {10z - 3{z^2} - 3} \right)}}} \frac{{dz}}{{iz}} = - \int\limits_{|z| = 1} {\frac{{{z^6} - 1}}{{{z^3}\left( {10z - 3{z^2} - 3} \right)}}} dz = \int\limits_{|z| = 1} {\frac{{{z^6} - 1}}{{{z^3}\left( {3{z^2} - 10z + 3} \right)}}} dz= \int\limits_{|z| = 1} {\frac{{{z^6} - 1}}{{3{z^3}\left( {z - 3} \right)\left( {z - \frac{1}{3}} \right)}}} dz$

Singular function points:

${z_1} = 0$ - third order pole

${z_2} = \frac{1}{3},\,\,{z_3} = 3$ - simple poles.

only points ${z_1} = 0$, ${z_2} = \frac{1}{3}$ fall inside the circle $|z|=1$ .

find the residues at these points:

$\mathop {res}\limits_{z = 0} f(z) = \frac{1}{{2!}}\mathop {\lim }\limits_{z \to 0} {\left( {\frac{{{z^6} - 1}}{{3{z^3}\left( {z - 3} \right)\left( {z - \frac{1}{3}} \right)}}{z^3}} \right)^{\prime \prime }} = \frac{1}{2}\mathop {\lim }\limits_{z \to 0} {\left( {\frac{{{z^6} - 1}}{{3{z^2} - 10z + 3}}} \right)^{\prime \prime }} = \frac{1}{2}\mathop {\lim }\limits_{z \to 0} {\left( {\frac{{6{z^5}\left( {3{z^2} - 10z + 3} \right) - \left( {{z^6} - 1} \right)\left( {6z - 10} \right)}}{{{{\left( {3{z^2} - 10z + 3} \right)}^2}}}} \right)^\prime } = \frac{1}{2}\mathop {\lim }\limits_{z \to 0} {\left( {\frac{{18{z^7} - 60{z^6} + 18{z^5} - 6{z^7} + 6z + 10{z^6} - 10}}{{{{\left( {3{z^2} - 10z + 3} \right)}^2}}}} \right)^\prime } = \frac{1}{2}\mathop {\lim }\limits_{z \to 0} {\left( {\frac{{12{z^7} - 50{z^6} + 18{z^5} + 6z - 10}}{{{{\left( {3{z^2} - 10z + 3} \right)}^2}}}} \right)^\prime } = \frac{1}{2}\mathop {\lim }\limits_{z \to 0} \frac{{\left( {84{z^6} - 300{z^5} + 90{z^4} + 6} \right){{\left( {3{z^2} - 10z + 3} \right)}^2} - \left( {12{z^7} - 50{z^6} + 18{z^5} + 6z - 10} \right) \cdot 2\left( {3{z^2} - 10z + 3} \right) \cdot \left( {6z - 10} \right)}}{{{{\left( {3{z^2} - 10z + 3} \right)}^4}}} = \frac{1}{2} \cdot \frac{{6 \cdot {3^2} + 10 \cdot 2 \cdot 3\left( { - 10} \right)}}{{{3^4}}} = \frac{{ - 91}}{{27}}$

$\mathop {res}\limits_{z = \frac{1}{3}} f(z) = \mathop {\lim }\limits_{z \to \frac{1}{3}} \frac{{{z^6} - 1}}{{3{z^3}\left( {z - 3} \right)\left( {z - \frac{1}{3}} \right)}}\left( {z - \frac{1}{3}} \right) = \mathop {\lim }\limits_{z \to \frac{1}{3}} \frac{{{z^6} - 1}}{{3{z^3}\left( {z - 3} \right)}} = \frac{{91}}{{27}}$

Then

$\int\limits_{|z| = 1} {\frac{{{z^6} - 1}}{{3{z^3}\left( {z - 3} \right)\left( {z - \frac{1}{3}} \right)}}} dz = 2\pi i\sum {res} f(z) = 2\pi i\left( { - \frac{{91}}{{27}} + \frac{{91}}{{27}}} \right) = 0$

Answer: 0