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Question Number 200130 by universe last updated on 14/Nov/23

$$solvebycontourintegrstion$$$${\int }_0^{2\pi }\frac{dx}{1+a\cos x}$$

Answered by Mathspace last updated on 14/Nov/23

$$I={\int }_0^{2\pi }\frac{dx}{1+a\frac{e^{ix}+e^{-ix}}{2}}(e^{ix}=z)$$$$={\int }_{\mid z\mid =1}\frac{dz}{iz(1+a\frac{z+z^{-1}}{2})}$$$$={\int }_{\mid z\mid =1}\frac{-2idz}{z(2+az+{az}^{-1})}$$$$={\int }_{\mid z\mid =1}\frac{-2idz}{2z+{az}^2+a}$$$$={\int }_{\mid z\mid =1}\frac{-2idz}{{az}^2+2z+a}$$$$f\left(z\right)=\frac{-2i}{{az}^2+2z+a}poles?$$$${\mathrm{\Delta }}^{{}^{\prime }}=1-a^2$$$$case1\mid a\mid <1\Rightarrow z_1=\frac{-1+\sqrt{1-a^2}}{a}$$$$z_2=\frac{-1-\sqrt{1-a^2}}{a}(a\ne 0)$$$$\mid z_1\mid -1=\frac{1-\sqrt{1-a^2}}{\mid a\mid }-1$$$$=\frac{1-\sqrt{1-a^2}-\mid a\mid }{\mid a\mid }=\frac{1-\mid a\mid -\sqrt{1-a^2}}{\mid a\mid }$$$${(1-\mid a\mid )}^2-(1-a^2)$$$$=1-2\mid a\mid +a^2-1+a^2$$$$=2a^2-2\mid a\mid =2\mid a\mid (\mid a\mid -1)<0$$$$\Rightarrow \mid z_1\mid <1$$$$\mid z_2\mid -1=\frac{1+\sqrt{1-a^2}}{\mid a\mid }-1$$$$=\frac{1-\mid a\mid +\sqrt{1-a^2}}{\mid a\mid }>0\Rightarrow \mid z_2\mid >1$$$$f\left(z\right)=\frac{-2i}{a(z-z_1)(z-z_2)}$$$${\int }_{\mid z\mid =1}f\left(z\right)dz=2i\pi Res(f,z_1)$$$$=2i\pi .\frac{-2i}{a(z_1-z_2)}=\frac{4\pi }{a\left(2\frac{\sqrt{1-a^2}}{a}\right)}$$$$=\frac{2\pi }{\sqrt{1-a^2}}\Rightarrow$$$$I=\frac{2\pi }{\sqrt{1-a^2}}$$$$restthecase\Rightarrow \mid a\mid >1$$$$z_1=\frac{-1+i\sqrt{a^2-1}}{a}$$$$and{z}_2=\frac{-1-i\sqrt{a^2-1}}{a}$$$$....followthesamepath...$$$$$$

Answered by MM42 last updated on 15/Nov/23

$$I=\int \frac{dx}{1+acosx}$$$$ifa=0\Rightarrow I=x+c$$$$ifa=1\Rightarrow I=\frac{1}{2}\int (1+{tan}^2\frac{x}{2})dx\Rightarrow I=tan\frac{x}{2}+c$$$$ifa=-1\Rightarrow I=\frac{1}{2}\int (1+{cotan}^2\frac{x}{2})dx\Rightarrow I=-cotan\frac{x}{2}+c$$$$otherwise$$$$\Rightarrow lettan\frac{x}{2}=u\Rightarrow \int \frac{2du}{(1+a)+(1-a)u^2}=\frac{2}{1-a}\int \frac{du}{\frac{1+a}{1-a}+u^2}$$$$let\frac{1+a}{1-a}=k\Rightarrow I=\frac{2}{1-a}\int \frac{du}{u^2+k}$$$$if\mid a\mid <1\Rightarrow k>0\Rightarrow I=\frac{2}{1-a}\times \frac{1}{\sqrt{k}}\times {tan}^{-1}\left(\frac{u}{\sqrt{k}}\right)+c$$$$=\frac{2}{\sqrt{1-a^2}}\times {tan}^{-1}\left(\sqrt{\frac{1-a}{1+a}}tan\left(\frac{x}{2}\times \sqrt{\frac{1-a}{1+a}}\right)\right)+c$$$$$$$$if\mid a\mid >1\Rightarrow k<0\Rightarrow I=\frac{1}{1-a}\times \frac{1}{k}\times ln\left(\frac{u-k}{u+k}\right)+c$$$$=\frac{1}{\sqrt{1-a^2}}\times ln\left(\frac{(1-a)tan\frac{x}{2}-\sqrt{1+a}}{(1-a)tan\frac{x}{2}+\sqrt{1+a}}\right)+c$$$$$$

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