3-cos-x-3-cos-x-dx-

Question Number 167989 by cortano1 last updated on 31/Mar/22

\int \frac{3 - \cos x}{3 + \cos x} d x = ?

Answered by nimnim last updated on 31/Mar/22

I = \int \frac{6 - 3 - cosx}{3 + cosx} dx = \int (\frac{6}{3 + cosx} - 1) dx

= 6 \int \frac{dx}{3 + cosx} - \int dx = I_{1} - x

put t = \tan (\frac{x}{2}) \Rightarrow \frac{2}{1 + t^{2}} dt = dx

sinx = \frac{2 t}{1 + t^{2}}, cosx = \frac{1 - t^{2}}{1 + t^{2}}

I_{1} = 6 \int \frac{1}{3 + \frac{1 - t^{2}}{1 + t^{2}}} \times \frac{2}{1 + t^{2}} dt

= 6 \int \frac{2}{4 + {2 t}^{2}} dt = 6 \sqrt{2} \tan^{- 1} (\frac{t}{\sqrt{2}})

= 6 \sqrt{2} \tan^{- 1} (\frac{1}{\sqrt{2}} \tan (\frac{x}{2}))

I = I_{1} - x = 6 \sqrt{2} \tan^{- 1} (\frac{1}{\sqrt{2}} \tan (\frac{x}{2})) - x + C

Answered by Florian last updated on 10/Apr/22

\int \frac{3 - c o s (x)}{3 + c o s (x)} d x = \int - 1 + \frac{6}{c o s (x) + 3} d x

= - \int 1 d x + \int \frac{6}{c o s (x) + 3} d x

= - x + 3 \sqrt{2} a r c t a n (\frac{1}{\sqrt{2}} t a n (\frac{x}{2})) + c, c \in R

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