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
$$\int\:\mathrm{cot}^{−\mathrm{1}} \left({x}\right)\:{dx}\:=\:\mathrm{I} \\ $$$$\mathrm{by}\:\mathrm{parts}\:.\:\mathrm{u}\:=\:\mathrm{cot}^{−\mathrm{1}} \left({x}\right)\Rightarrow{du}=−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{dx} \\ $$$${dv}\:=\:{dx}\:\Rightarrow\:{v}\:=\:{x} \\ $$$$\mathrm{I}\:=\:{x}\:\mathrm{cot}^{−\mathrm{1}} \left({x}\right)+\:\int\:{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{dx}\: \\ $$$$\mathrm{I}\:=\:{x}\:\mathrm{cot}^{−\mathrm{1}} \left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}\int\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{d}\left(\mathrm{1}−{x}^{\mathrm{2}} \right) \\ $$$$\mathrm{I}\:=\:{x}\:\mathrm{cot}^{−\mathrm{1}} \left({x}\right)−\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:+\:{c} \\ $$
Commented by Kunal12588 last updated on 30/May/20
![(1/(tan^(−1) x)) ≠ cot^(−1) x [tan^(−1) ((1/x))=cot^(−1) x]](https://www.tinkutara.com/question/Q96187.png)
$$\frac{\mathrm{1}}{{tan}^{−\mathrm{1}} {x}}\:\neq\:{cot}^{−\mathrm{1}} {x} \\ $$$$\left[{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\right)={cot}^{−\mathrm{1}} {x}\right] \\ $$
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
$$\mathrm{this}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{solved} \\ $$