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Question Number 156305 by cortano last updated on 10/Oct/21

  ∫ ((sec^2 (x))/(3sin^2 (x)−1)) dx

$$\:\:\int\:\frac{\mathrm{sec}\:^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{3sin}\:^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{1}}\:\mathrm{dx} \\ $$

Answered by puissant last updated on 10/Oct/21

Ω=∫((sec^2 x)/(3sin^2 x−1))dx = ∫((sec^4 x)/(3tan^2 x−sec^2 x))dx  u=tanx → du=sec^2 xdx=(1+u^2 )du  ⇒ Ω=∫((1+u^2 )/(3u^2 −(1+u^2 )))du=∫((1+u^2 )/(2u^2 −1))du  Ω=∫(1/(2u^2 −1))du+∫(u^2 /(2u^2 −1))du  =(1/2)∫(du/(u^2 −(1/2)))+(1/2)∫((2u^2 −1+1)/(2u^2 −1))du  =(1/2)∫(du/(u^2 −(1/2)))+(1/4)∫(du/(u^2 −(1/2)))+(1/2)∫du =(3/4)∫(du/(u^2 −(1/2)))+(1/2)∫du  =−(6/4)∫(du/(1−((u/( (√2))))))−(u/2) = −((6(√2))/4) argtanh((√2)u)−(u/2)+C..    ∴∵  Ω = −((3(√2))/2) argtanh((√2)tanx)−((tanx)/2)+C

$$\Omega=\int\frac{{sec}^{\mathrm{2}} {x}}{\mathrm{3}{sin}^{\mathrm{2}} {x}−\mathrm{1}}{dx}\:=\:\int\frac{{sec}^{\mathrm{4}} {x}}{\mathrm{3}{tan}^{\mathrm{2}} {x}−{sec}^{\mathrm{2}} {x}}{dx} \\ $$$${u}={tanx}\:\rightarrow\:{du}={sec}^{\mathrm{2}} {xdx}=\left(\mathrm{1}+{u}^{\mathrm{2}} \right){du} \\ $$$$\Rightarrow\:\Omega=\int\frac{\mathrm{1}+{u}^{\mathrm{2}} }{\mathrm{3}{u}^{\mathrm{2}} −\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}{du}=\int\frac{\mathrm{1}+{u}^{\mathrm{2}} }{\mathrm{2}{u}^{\mathrm{2}} −\mathrm{1}}{du} \\ $$$$\Omega=\int\frac{\mathrm{1}}{\mathrm{2}{u}^{\mathrm{2}} −\mathrm{1}}{du}+\int\frac{{u}^{\mathrm{2}} }{\mathrm{2}{u}^{\mathrm{2}} −\mathrm{1}}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{du}}{{u}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{u}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}}{\mathrm{2}{u}^{\mathrm{2}} −\mathrm{1}}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{du}}{{u}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{du}}{{u}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{2}}\int{du}\:=\frac{\mathrm{3}}{\mathrm{4}}\int\frac{{du}}{{u}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{2}}\int{du} \\ $$$$=−\frac{\mathrm{6}}{\mathrm{4}}\int\frac{{du}}{\mathrm{1}−\left(\frac{{u}}{\:\sqrt{\mathrm{2}}}\right)}−\frac{{u}}{\mathrm{2}}\:=\:−\frac{\mathrm{6}\sqrt{\mathrm{2}}}{\mathrm{4}}\:{argtanh}\left(\sqrt{\mathrm{2}}{u}\right)−\frac{{u}}{\mathrm{2}}+{C}.. \\ $$$$ \\ $$$$\therefore\because\:\:\Omega\:=\:−\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{2}}\:{argtanh}\left(\sqrt{\mathrm{2}}{tanx}\right)−\frac{{tanx}}{\mathrm{2}}+{C} \\ $$

Commented by cortano last updated on 10/Oct/21

ok. i got in tan x form.

$$\mathrm{ok}.\:\mathrm{i}\:\mathrm{got}\:\mathrm{in}\:\mathrm{tan}\:\mathrm{x}\:\mathrm{form}. \\ $$

Commented by puissant last updated on 10/Oct/21

Okey...

$${Okey}... \\ $$

Commented by puissant last updated on 10/Oct/21

Mr Cortano , another important   formula :  argth(x)+C=(1/2)ln(((x+1)/(x−1)))+C=∫(1/(x^2 −1))dx..

$${Mr}\:{Cortano}\:,\:{another}\:{important}\: \\ $$$${formula}\:: \\ $$$${argth}\left({x}\right)+{C}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)+{C}=\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}{dx}.. \\ $$

Commented by cortano last updated on 10/Oct/21

thank you sir

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

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