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bemath-2-cos-x-2-cos-x-dx-




Question Number 111873 by bemath last updated on 05/Sep/20
   (√(bemath))  ∫ ((2−cos x)/(2+cos x)) dx
$$\:\:\:\sqrt{{bemath}} \\ $$$$\int\:\frac{\mathrm{2}−\mathrm{cos}\:{x}}{\mathrm{2}+\mathrm{cos}\:{x}}\:{dx}\: \\ $$
Answered by Lordose last updated on 05/Sep/20
  Set u=tan(x/2) ⇒du=(1/2)sec^2 (x/2)dx  2∫((2−(((1−u^2 )/(1+u^2 ))))/((1+u^2 )(((1−u^2 )/(1+u^2 )) + 2)))du  2∫(((((2+2u^2 −1+u^2 )/(1+u^2 ))))/(1−u^2 +2+2u^2 ))du  2∫(((1+3u^2 )/(1+u^2 )) × (1/(3+u^2 )))du  2∫((1+3u^2 )/(u^4 +4u^2 +3))du  Resolving into P.F  2∫((4/(u^2 +3)) − (1/(u^2 +1)))du  8∫(1/(u^2 +3))du − 2∫(1/(u^2 +1))du  (8/3)∫((1/((u^2 /3)+1)))du − 2(tan^(−1) u)  set y=(u/( (√3)))  ⇒dy=(du/( (√3)))  ((8(√3))/3)∫(1/(y^2 +1))dy − 2tan^(−1) (tan(x/2))  ((8(√3))/3)(tan^(−1) y) − x + C  ((8(√3))/3)(tan^(−1) ((u/( (√3))))) − x +C  ((8(√3))/3)(tan^(−1) (((tan(x/2))/( (√3))))) − x +C  ★LorD OsE
$$ \\ $$$$\boldsymbol{\mathrm{Set}}\:\boldsymbol{\mathrm{u}}=\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\:\Rightarrow\boldsymbol{\mathrm{du}}=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{sec}}^{\mathrm{2}} \frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\boldsymbol{\mathrm{dx}} \\ $$$$\mathrm{2}\int\frac{\mathrm{2}−\left(\frac{\mathrm{1}−\boldsymbol{\mathrm{u}}^{\mathrm{2}} }{\mathrm{1}+\boldsymbol{\mathrm{u}}^{\mathrm{2}} }\right)}{\left(\mathrm{1}+\boldsymbol{\mathrm{u}}^{\mathrm{2}} \right)\left(\frac{\mathrm{1}−\boldsymbol{\mathrm{u}}^{\mathrm{2}} }{\mathrm{1}+\boldsymbol{\mathrm{u}}^{\mathrm{2}} }\:+\:\mathrm{2}\right)}\boldsymbol{\mathrm{du}} \\ $$$$\mathrm{2}\int\frac{\left(\frac{\mathrm{2}+\mathrm{2}\boldsymbol{\mathrm{u}}^{\mathrm{2}} −\mathrm{1}+\boldsymbol{\mathrm{u}}^{\mathrm{2}} }{\mathrm{1}+\boldsymbol{\mathrm{u}}^{\mathrm{2}} }\right)}{\mathrm{1}−\boldsymbol{\mathrm{u}}^{\mathrm{2}} +\mathrm{2}+\mathrm{2}\boldsymbol{\mathrm{u}}^{\mathrm{2}} }\boldsymbol{\mathrm{du}} \\ $$$$\mathrm{2}\int\left(\frac{\mathrm{1}+\mathrm{3}\boldsymbol{\mathrm{u}}^{\mathrm{2}} }{\mathrm{1}+\boldsymbol{\mathrm{u}}^{\mathrm{2}} }\:×\:\frac{\mathrm{1}}{\mathrm{3}+\boldsymbol{\mathrm{u}}^{\mathrm{2}} }\right)\boldsymbol{\mathrm{du}} \\ $$$$\mathrm{2}\int\frac{\mathrm{1}+\mathrm{3}\boldsymbol{\mathrm{u}}^{\mathrm{2}} }{\boldsymbol{\mathrm{u}}^{\mathrm{4}} +\mathrm{4}\boldsymbol{\mathrm{u}}^{\mathrm{2}} +\mathrm{3}}\boldsymbol{\mathrm{du}} \\ $$$$\boldsymbol{\mathrm{Resolving}}\:\boldsymbol{\mathrm{into}}\:\boldsymbol{\mathrm{P}}.\boldsymbol{\mathrm{F}} \\ $$$$\mathrm{2}\int\left(\frac{\mathrm{4}}{\boldsymbol{\mathrm{u}}^{\mathrm{2}} +\mathrm{3}}\:−\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{u}}^{\mathrm{2}} +\mathrm{1}}\right)\boldsymbol{\mathrm{du}} \\ $$$$\mathrm{8}\int\frac{\mathrm{1}}{\boldsymbol{\mathrm{u}}^{\mathrm{2}} +\mathrm{3}}\boldsymbol{\mathrm{du}}\:−\:\mathrm{2}\int\frac{\mathrm{1}}{\boldsymbol{\mathrm{u}}^{\mathrm{2}} +\mathrm{1}}\boldsymbol{\mathrm{du}} \\ $$$$\frac{\mathrm{8}}{\mathrm{3}}\int\left(\frac{\mathrm{1}}{\frac{\boldsymbol{\mathrm{u}}^{\mathrm{2}} }{\mathrm{3}}+\mathrm{1}}\right)\boldsymbol{\mathrm{du}}\:−\:\mathrm{2}\left(\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \boldsymbol{\mathrm{u}}\right) \\ $$$$\boldsymbol{\mathrm{set}}\:\boldsymbol{\mathrm{y}}=\frac{\boldsymbol{\mathrm{u}}}{\:\sqrt{\mathrm{3}}}\:\:\Rightarrow\boldsymbol{\mathrm{dy}}=\frac{\boldsymbol{\mathrm{du}}}{\:\sqrt{\mathrm{3}}} \\ $$$$\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{3}}\int\frac{\mathrm{1}}{\boldsymbol{\mathrm{y}}^{\mathrm{2}} +\mathrm{1}}\boldsymbol{\mathrm{dy}}\:−\:\mathrm{2}\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \left(\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \boldsymbol{\mathrm{y}}\right)\:−\:\boldsymbol{\mathrm{x}}\:+\:\boldsymbol{\mathrm{C}} \\ $$$$\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \left(\frac{\boldsymbol{\mathrm{u}}}{\:\sqrt{\mathrm{3}}}\right)\right)\:−\:\boldsymbol{\mathrm{x}}\:+\boldsymbol{\mathrm{C}} \\ $$$$\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \left(\frac{\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\right)\right)\:−\:\boldsymbol{\mathrm{x}}\:+\boldsymbol{\mathrm{C}} \\ $$$$\bigstar\boldsymbol{\mathrm{LorD}}\:\boldsymbol{\mathrm{OsE}} \\ $$
Commented by bemath last updated on 05/Sep/20
jooss sir
$${jooss}\:{sir} \\ $$$$ \\ $$

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