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

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

\sqrt{b e m a t h}

\int \frac{2 - \cos x}{2 + \cos x} d x

Answered by Lordose last updated on 05/Sep/20

Set u = \tan \frac{x}{2} \Rightarrow du = \frac{1}{2} \sec^{2} \frac{x}{2} dx

2 \int \frac{2 - (\frac{1 - u^{2}}{1 + u^{2}})}{(1 + u^{2}) (\frac{1 - u^{2}}{1 + u^{2}} + 2)} du

2 \int \frac{(\frac{2 + 2 u^{2} - 1 + u^{2}}{1 + u^{2}})}{1 - u^{2} + 2 + 2 u^{2}} du

2 \int (\frac{1 + 3 u^{2}}{1 + u^{2}} \times \frac{1}{3 + u^{2}}) du

2 \int \frac{1 + 3 u^{2}}{u^{4} + 4 u^{2} + 3} du

Resolving into P . F

2 \int (\frac{4}{u^{2} + 3} - \frac{1}{u^{2} + 1}) du

8 \int \frac{1}{u^{2} + 3} du - 2 \int \frac{1}{u^{2} + 1} du

\frac{8}{3} \int (\frac{1}{\frac{u^{2}}{3} + 1}) du - 2 (\tan^{- 1} u)

set y = \frac{u}{\sqrt{3}} \Rightarrow dy = \frac{du}{\sqrt{3}}

\frac{8 \sqrt{3}}{3} \int \frac{1}{y^{2} + 1} dy - 2 \tan^{- 1} (\tan \frac{x}{2})

\frac{8 \sqrt{3}}{3} (\tan^{- 1} y) - x + C

\frac{8 \sqrt{3}}{3} (\tan^{- 1} (\frac{u}{\sqrt{3}})) - x + C

\frac{8 \sqrt{3}}{3} (\tan^{- 1} (\frac{\tan \frac{x}{2}}{\sqrt{3}})) - x + C

★ LorD OsE

Commented by bemath last updated on 05/Sep/20

j o o s s s i r

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