Question Number 20666 by NECx last updated on 31/Aug/17
$$\int\mathrm{cot}\:^{\mathrm{4}} {xdx} \\ $$
Answered by Joel577 last updated on 31/Aug/17
$$\mathrm{cosec}^{\mathrm{2}} \:{x}\:−\:\mathrm{cot}^{\mathrm{2}} \:{x}\:=\:\mathrm{1} \\ $$$${I}\:=\:\int\:\mathrm{cot}^{\mathrm{2}} \:{x}\left(\mathrm{cosec}^{\mathrm{2}} \:{x}\:−\:\mathrm{1}\right)\:{dx} \\ $$$$\:\:\:=\:\int\:\left(\mathrm{cot}^{\mathrm{2}} \:{x}\:\mathrm{cosec}^{\mathrm{2}} \:{x}\right)\:{dx}\:−\:\int\:\mathrm{cot}^{\mathrm{2}} \:{x}\:{dx} \\ $$$$\mathrm{Let}\:{u}\:=\:\mathrm{cot}\:{x}\:\:\rightarrow\:\:{du}\:=\:−\mathrm{cosec}^{\mathrm{2}} \:{x}\:{dx} \\ $$$${I}\:=\:\int\:{u}^{\mathrm{2}} \:.\:\mathrm{cosec}^{\mathrm{2}} \:{x}\:.\:\frac{{du}}{−\mathrm{cosec}^{\mathrm{2}} \:{x}}\:−\:\int\:\left(\mathrm{cosec}^{\mathrm{2}} \:{x}\:−\:\mathrm{1}\right)\:{dx} \\ $$$$\:\:\:=\:−\frac{\mathrm{1}}{\mathrm{3}}{u}^{\mathrm{3}} \:+\:\mathrm{cot}\:{x}\:+\:{x}\:+\:{C} \\ $$$$\:\:\:=\:−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{cot}^{\mathrm{3}} \:{x}\:+\:\mathrm{cot}\:{x}\:+\:{x}\:+\:{C} \\ $$
Commented by NECx last updated on 31/Aug/17
$${thank}\:{you}\:{sir}. \\ $$