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

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

Answered by nimnim last updated on 31/Mar/22

$$\mathrm{I}=\int \frac{6-3-\mathrm{cosx}}{3+\mathrm{cosx}}\mathrm{dx}=\int (\frac{6}{3+\mathrm{cosx}}-1)\mathrm{dx}$$$$=6\int \frac{\mathrm{dx}}{3+\mathrm{cosx}}-\int \mathrm{dx}={\mathrm{I}}_1-\mathrm{x}$$$$\mathrm{put}\mathrm{t}=\tan \left(\frac{\mathrm{x}}{2}\right)\Rightarrow \frac{2}{1+{\mathrm{t}}^2}\mathrm{dt}=\mathrm{dx}$$$$\mathrm{sinx}=\frac{2\mathrm{t}}{1+{\mathrm{t}}^2},\mathrm{cosx}=\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}$$$${\mathrm{I}}_1=6\int \frac{1}{3+\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}}\times \frac{2}{1+{\mathrm{t}}^2}\mathrm{dt}$$$$=6\int \frac{2}{4+{2\mathrm{t}}^2}\mathrm{dt}=6\sqrt{2}{\tan }^{-1}\left(\frac{\mathrm{t}}{\sqrt{2}}\right)$$$$=6\sqrt{2}{\tan }^{-1}\left(\frac{1}{\sqrt{2}}\tan \left(\frac{\mathrm{x}}{2}\right)\right)$$$$\mathrm{I}={\mathrm{I}}_1-\mathrm{x}=6\sqrt{2}{\tan }^{-1}\left(\frac{1}{\sqrt{2}}\tan \left(\frac{\mathrm{x}}{2}\right)\right)-\mathrm{x}+\mathrm{C}$$

Answered by Florian last updated on 10/Apr/22

$$\int \frac{3-cos\left(x\right)}{3+cos\left(x\right)}dx=\int -1+\frac{6}{cos\left(x\right)+3}dx$$$$=-\int 1dx+\int \frac{6}{cos\left(x\right)+3}dx$$$$=-x+3\sqrt{2}arctan\left(\frac{1}{\sqrt{2}}tan\left(\frac{x}{2}\right)\right)+c,c\in R$$

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