find-dx-tan-1-x-

Question Number 96092 by mhmd last updated on 29/May/20

$${find}\:\int\frac{{dx}}{{tan}^{−\mathrm{1}} \left({x}\right)} \\ $$

Commented by bemath last updated on 30/May/20

∫ cot^(−1) (x) dx = I by parts . u = cot^(−1) (x)⇒du=−(√(1−x^2 )) dx dv = dx ⇒ v = x I = x cot^(−1) (x)+ ∫ x(√(1−x^2 )) dx I = x cot^(−1) (x)−(1/2)∫ (√(1−x^2 )) d(1−x^2 ) I = x cot^(−1) (x)−(1/3)(√((1−x^2 )^3 )) + c

$$\int\:\mathrm{cot}^{−\mathrm{1}} \left({x}\right)\:{dx}\:=\:\mathrm{I} \\ $$$$\mathrm{by}\:\mathrm{parts}\:.\:\mathrm{u}\:=\:\mathrm{cot}^{−\mathrm{1}} \left({x}\right)\Rightarrow{du}=−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{dx} \\ $$$${dv}\:=\:{dx}\:\Rightarrow\:{v}\:=\:{x} \\ $$$$\mathrm{I}\:=\:{x}\:\mathrm{cot}^{−\mathrm{1}} \left({x}\right)+\:\int\:{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{dx}\: \\ $$$$\mathrm{I}\:=\:{x}\:\mathrm{cot}^{−\mathrm{1}} \left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}\int\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{d}\left(\mathrm{1}−{x}^{\mathrm{2}} \right) \\ $$$$\mathrm{I}\:=\:{x}\:\mathrm{cot}^{−\mathrm{1}} \left({x}\right)−\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:+\:{c} \\ $$

Commented by Kunal12588 last updated on 30/May/20

(1/(tan^(−1) x)) ≠ cot^(−1) x [tan^(−1) ((1/x))=cot^(−1) x]

$$\frac{\mathrm{1}}{{tan}^{−\mathrm{1}} {x}}\:\neq\:{cot}^{−\mathrm{1}} {x} \\ $$$$\left[{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\right)={cot}^{−\mathrm{1}} {x}\right] \\ $$

Commented by mathmax by abdo last updated on 30/May/20

$$\mathrm{answer}\:\mathrm{not}\:\mathrm{correct}! \\ $$

Commented by MJS last updated on 30/May/20

$$\mathrm{this}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{solved} \\ $$

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