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Question Number 20666 by NECx last updated on 31/Aug/17

∫cot^4 xdx

$$\int\mathrm{cot}\:^{\mathrm{4}} {xdx} \\ $$

Answered by Joel577 last updated on 31/Aug/17

cosec^2 x − cot^2 x = 1 I = ∫ cot^2 x(cosec^2 x − 1) dx = ∫ (cot^2 x cosec^2 x) dx − ∫ cot^2 x dx Let u = cot x → du = −cosec^2 x dx I = ∫ u^2 . cosec^2 x . (du/(−cosec^2 x)) − ∫ (cosec^2 x − 1) dx = −(1/3)u^3 + cot x + x + C = −(1/3)cot^3 x + cot x + x + C

$$\mathrm{cosec}^{\mathrm{2}} \:{x}\:−\:\mathrm{cot}^{\mathrm{2}} \:{x}\:=\:\mathrm{1} \\ $$$${I}\:=\:\int\:\mathrm{cot}^{\mathrm{2}} \:{x}\left(\mathrm{cosec}^{\mathrm{2}} \:{x}\:−\:\mathrm{1}\right)\:{dx} \\ $$$$\:\:\:=\:\int\:\left(\mathrm{cot}^{\mathrm{2}} \:{x}\:\mathrm{cosec}^{\mathrm{2}} \:{x}\right)\:{dx}\:−\:\int\:\mathrm{cot}^{\mathrm{2}} \:{x}\:{dx} \\ $$$$\mathrm{Let}\:{u}\:=\:\mathrm{cot}\:{x}\:\:\rightarrow\:\:{du}\:=\:−\mathrm{cosec}^{\mathrm{2}} \:{x}\:{dx} \\ $$$${I}\:=\:\int\:{u}^{\mathrm{2}} \:.\:\mathrm{cosec}^{\mathrm{2}} \:{x}\:.\:\frac{{du}}{−\mathrm{cosec}^{\mathrm{2}} \:{x}}\:−\:\int\:\left(\mathrm{cosec}^{\mathrm{2}} \:{x}\:−\:\mathrm{1}\right)\:{dx} \\ $$$$\:\:\:=\:−\frac{\mathrm{1}}{\mathrm{3}}{u}^{\mathrm{3}} \:+\:\mathrm{cot}\:{x}\:+\:{x}\:+\:{C} \\ $$$$\:\:\:=\:−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{cot}^{\mathrm{3}} \:{x}\:+\:\mathrm{cot}\:{x}\:+\:{x}\:+\:{C} \\ $$

Commented by NECx last updated on 31/Aug/17

thank you sir.

$${thank}\:{you}\:{sir}. \\ $$

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