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