Put $\cos ecx + \cot x = z$
$\cos ec - \cot x = \frac{1}{z}.$
$ - 2\cos e{c^2}xdx = \left( {1 + \frac{1}{{{z^2}}}} \right)dz$
$\therefore I = - \frac{1}{2}\int {\frac{{1 + \frac{1}{{{z^2}}}}}{{{z^{9/2}}}}} dz$
$ = \frac{1}{2}\left[ {\int {{z^{ - 9/2}}dz + \int {{z^{\frac{{ - 13}}{2}}}} dz} } \right]$
$ = - \frac{1}{2}\left[ {\frac{{{z^{ - 7/2}}}}{{\left( { - 7} \right)}}2 + \frac{{{z^{ - 11/2}}}}{{\left( { - 11} \right)}}2} \right] + C$
$ = {z^{\frac{{ - 7}}{2}}}\left[ {\frac{1}{7} + \frac{{{z^{ - 3}}}}{{11}}} \right] + C$
$ = {\left( {\cos ecx - \cot x} \right)^{\frac{7}{2}}}\left( {\frac{1}{7} + \frac{{{{\left( {\cos ecx - \cot x} \right)}^2}}}{{11}}} \right) + C$
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If $I_1 = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} \, f (\tan\, \theta + \cot\, \theta )\cdot sec^2\, \theta\, d\, \theta$ &
$I_2 = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} \, f (\tan\, \theta + \cot\, \theta )\cdot cosec^2\, \theta\, d \, \theta$ ,
then the ratio $\frac{{{I_1}}}{{{I_2}}}$ :