Question Number 219660 by Nicholas666 last updated on 30/Apr/25
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\int_{\:\mathrm{0}} ^{\:\mathrm{2}\pi} \frac{\mathrm{1}}{{a}\:+\:{b}\:{cos}\:\left({x}\right)}\:{dx} \\ $$$$ \\ $$
Answered by vnm last updated on 01/May/25
![if ∣a∣<∣b∣ the integral diverges if ∣a∣>∣b∣: ∫(1/(a+bcosx))dx=[tan(x/2)=u, x=2tan^(−1) u, dx=((2du)/(1+u^2 ))]= sgna∫(1/(∣a∣+sgna∙b((1−u^2 )/(1+u^2 ))))((2du)/(1+u^2 ))=2sgna∫(du/(∣a∣(1+u^2 )+sgna∙b(1−u^2 )))= 2sgna∫(du/((∣a∣+sgna∙b)+(∣a∣−sgna∙b)u^2 ))=((2sgna)/(a−b))∫(du/(u^2 +((√((a+b)/(a−b))))^2 ))= ((2sgna)/(a−b))∙(√((a−b)/(a+b)))tan^(−1) (u(√((a−b)/(a+b))))+C=((2sgna)/( (√(a^2 −b^2 ))))tan^(−1) (tan(x/2)∙(√((a−b)/(a+b))))+C ∫_0 ^(2π) (1/(a+bcosx))dx=((2sgna)/( (√(a^2 −b^2 ))))tan^(−1) (tan(x/2)(√((a−b)/(a+b))))∣_0 ^(2π) =((2sgna)/( (√(a^2 −b^2 ))))(lim_(θ→0^+ ) tan^(−1) (tan(x/2)∙(√((a−b)/(a+b))))∣_0 ^(π−θ) +lim_(θ→0^+ ) tan^(−1) (tan(x/2)(√((a−b)/(a+b))))∣_(π+θ) ^(2π) )= ((2sgna)/( (√(a^2 −b^2 ))))(((π/2)−0)+(0−(−(π/2))))=((2πsgna)/( (√(a^2 −b^2 ))))](https://www.tinkutara.com/question/Q219706.png)
$$\mathrm{if}\:\mid{a}\mid<\mid{b}\mid\:\mathrm{the}\:\mathrm{integral}\:\mathrm{diverges} \\ $$$$\mathrm{if}\:\mid{a}\mid>\mid{b}\mid: \\ $$$$\int\frac{\mathrm{1}}{{a}+{b}\mathrm{cos}{x}}{dx}=\left[\mathrm{tan}\frac{{x}}{\mathrm{2}}={u},\:{x}=\mathrm{2tan}^{−\mathrm{1}} {u},\:\:{dx}=\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\right]= \\ $$$$\mathrm{sgn}{a}\int\frac{\mathrm{1}}{\mid{a}\mid+\mathrm{sgn}{a}\centerdot{b}\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }=\mathrm{2sgn}{a}\int\frac{{du}}{\mid{a}\mid\left(\mathrm{1}+{u}^{\mathrm{2}} \right)+\mathrm{sgn}{a}\centerdot{b}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}= \\ $$$$\mathrm{2sgn}{a}\int\frac{{du}}{\left(\mid{a}\mid+\mathrm{sgn}{a}\centerdot{b}\right)+\left(\mid{a}\mid−\mathrm{sgn}{a}\centerdot{b}\right){u}^{\mathrm{2}} }=\frac{\mathrm{2sgn}{a}}{{a}−{b}}\int\frac{{du}}{{u}^{\mathrm{2}} +\left(\sqrt{\frac{{a}+{b}}{{a}−{b}}}\right)^{\mathrm{2}} }= \\ $$$$\frac{\mathrm{2sgn}{a}}{{a}−{b}}\centerdot\sqrt{\frac{{a}−{b}}{{a}+{b}}}\mathrm{tan}^{−\mathrm{1}} \left({u}\sqrt{\frac{{a}−{b}}{{a}+{b}}}\right)+{C}=\frac{\mathrm{2sgn}{a}}{\:\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{tan}\frac{{x}}{\mathrm{2}}\centerdot\sqrt{\frac{{a}−{b}}{{a}+{b}}}\right)+{C} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{\mathrm{1}}{{a}+{b}\mathrm{cos}{x}}{dx}=\frac{\mathrm{2sgn}{a}}{\:\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{tan}\frac{{x}}{\mathrm{2}}\sqrt{\frac{{a}−{b}}{{a}+{b}}}\right)\mid_{\mathrm{0}} ^{\mathrm{2}\pi} =\frac{\mathrm{2sgn}{a}}{\:\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\left(\underset{\theta\rightarrow\mathrm{0}^{+} } {\mathrm{lim}tan}^{−\mathrm{1}} \left(\mathrm{tan}\frac{{x}}{\mathrm{2}}\centerdot\sqrt{\frac{{a}−{b}}{{a}+{b}}}\right)\mid_{\mathrm{0}} ^{\pi−\theta} +\underset{\theta\rightarrow\mathrm{0}^{+} } {\mathrm{lim}tan}^{−\mathrm{1}} \left(\mathrm{tan}\frac{{x}}{\mathrm{2}}\sqrt{\frac{{a}−{b}}{{a}+{b}}}\right)\mid_{\pi+\theta} ^{\mathrm{2}\pi} \right)= \\ $$$$\frac{\mathrm{2sgn}{a}}{\:\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\left(\left(\frac{\pi}{\mathrm{2}}−\mathrm{0}\right)+\left(\mathrm{0}−\left(−\frac{\pi}{\mathrm{2}}\right)\right)\right)=\frac{\mathrm{2}\pi\mathrm{sgn}{a}}{\:\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }} \\ $$