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Question Number 111873 by bemath last updated on 05/Sep/20

$$\sqrt{bemath}$$$$\int \frac{2-\cos x}{2+\cos x}dx$$

Answered by Lordose last updated on 05/Sep/20

$$$$$$\mathbf{Set}\mathbf{u}=\mathbf{tan}\frac{\mathbf{x}}{2}\Rightarrow \mathbf{du}=\frac{1}{2}{\mathbf{sec}}^2\frac{\mathbf{x}}{2}\mathbf{dx}$$$$2\int \frac{2-\left(\frac{1-{\mathbf{u}}^2}{1+{\mathbf{u}}^2}\right)}{(1+{\mathbf{u}}^2)(\frac{1-{\mathbf{u}}^2}{1+{\mathbf{u}}^2}+2)}\mathbf{du}$$$$2\int \frac{\left(\frac{2+2{\mathbf{u}}^2-1+{\mathbf{u}}^2}{1+{\mathbf{u}}^2}\right)}{1-{\mathbf{u}}^2+2+2{\mathbf{u}}^2}\mathbf{du}$$$$2\int \left(\frac{1+3{\mathbf{u}}^2}{1+{\mathbf{u}}^2}\times \frac{1}{3+{\mathbf{u}}^2}\right)\mathbf{du}$$$$2\int \frac{1+3{\mathbf{u}}^2}{{\mathbf{u}}^4+4{\mathbf{u}}^2+3}\mathbf{du}$$$$\mathbf{Resolving}\mathbf{into}\mathbf{P}.\mathbf{F}$$$$2\int (\frac{4}{{\mathbf{u}}^2+3}-\frac{1}{{\mathbf{u}}^2+1})\mathbf{du}$$$$8\int \frac{1}{{\mathbf{u}}^2+3}\mathbf{du}-2\int \frac{1}{{\mathbf{u}}^2+1}\mathbf{du}$$$$\frac{8}{3}\int \left(\frac{1}{\frac{{\mathbf{u}}^2}{3}+1}\right)\mathbf{du}-2\left({\mathbf{tan}}^{-1}\mathbf{u}\right)$$$$\mathbf{set}\mathbf{y}=\frac{\mathbf{u}}{\sqrt{3}}\Rightarrow \mathbf{dy}=\frac{\mathbf{du}}{\sqrt{3}}$$$$\frac{8\sqrt{3}}{3}\int \frac{1}{{\mathbf{y}}^2+1}\mathbf{dy}-2{\mathbf{tan}}^{-1}\left(\mathbf{tan}\frac{\mathbf{x}}{2}\right)$$$$\frac{8\sqrt{3}}{3}\left({\mathbf{tan}}^{-1}\mathbf{y}\right)-\mathbf{x}+\mathbf{C}$$$$\frac{8\sqrt{3}}{3}\left({\mathbf{tan}}^{-1}\left(\frac{\mathbf{u}}{\sqrt{3}}\right)\right)-\mathbf{x}+\mathbf{C}$$$$\frac{8\sqrt{3}}{3}\left({\mathbf{tan}}^{-1}\left(\frac{\mathbf{tan}\frac{\mathbf{x}}{2}}{\sqrt{3}}\right)\right)-\mathbf{x}+\mathbf{C}$$$$★\mathbf{LorD}\mathbf{OsE}$$

Commented by bemath last updated on 05/Sep/20

$$joosssir$$$$$$

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