Question

∫ cot³ 2x dx

Answer 1

let u=2x$\implies$ du=2dx

$\therefore∴ \int cot^32xdx= \frac12 \int cot^3udu...(i)$

let $\Iota_ n= \frac{ (-cotn-1x)} {( n-1)} -\Iota _{n-2}$

$\Iota _ {3}=\frac {(- cot 2u)} 2 - \Iota _1$

$= \frac {(- cot2 u)} 2- \int cot udu$

$= \frac {(- cot2u)} 2 - \Iota_n sin u+ c...(ii)$

(ii)and (i) gives

$\therefore∴ \int cot^3 2xdx= \frac {(-cot ^32x)}4 - \frac 12 \iota _nsin 2x + c$