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Question Number 46850 by maxmathsup by imad last updated on 01/Nov/18

$$let{a}^2>b^2+c^{2}calculate{\int }_0^{2\pi }\frac{d\theta }{a+bsin\theta +ccos\theta }$$

Commented bymaxmathsup by imad last updated on 01/Nov/18

$$residusmethodletI={\int }_0^{2\pi }\frac{d\theta }{a+bsin\theta +ccos\theta }changementz=e^{i\theta }give$$ $$I={\int }_{\mid z\mid =1}\frac{1}{a+b\frac{z-z^{-1}}{2i}+c\frac{z+z^{-1}}{2}}\frac{dz}{iz}={\int }_{\mid z\mid =1}\frac{dz}{iz\{a+b\frac{z-z^{-1}}{2i}+c\frac{z+z^{-1}}{2}\}}$$ $$={\int }_{\mid z\mid =1}\frac{2dz}{z\{2ia+b(z-z^{-1})+ci(z+z^{-1})\}}$$ $$={\int }_{\mid z\mid =1}\frac{2dz}{2iaz+{bz}^2-b+{ciz}^2+ci}={\int }_{\mid z\mid =1}\frac{2dz}{(b+ci)z^2+2iaz+ci-b}$$ $$let\phi \left(z\right)=\frac{2}{(b+ci)z^2+2iaz+ci-b}.polesof\phi ?$$ $${\mathrm{\Delta }}^{{}^{\prime }}={\left(ia\right)}^2-(b+ci)(ci-b)=-a^2-({\left(ci\right)}^2-b^2)$$ $$=-a^2+b^2+c^2=-(a^2-(b^2+c^2))<0\Rightarrow z_1=\frac{-ia+i\sqrt{a^2-b^2-c^2}}{b+ci}$$ $$z_2=\frac{-ia-i\sqrt{a^2-b^2-c^2}}{b+ci}$$ $$\mid z_1\mid -1=\frac{\mid a-\sqrt{a^2-b^2-c^2}\mid }{\sqrt{b^2+c^2}}>1\Rightarrow z_{1}isoutofcircle.\Rightarrow Res(\phi ,z_1)=0$$ $$\mid z_2\mid -1=\frac{\mid a+\sqrt{a^2-b^2-c^2}\mid }{\sqrt{b^2+c^2}}>1\Rightarrow z_{2}isoutofcircle\Rightarrow Res(\phi ,z_2)=0$$ $${\int }_0^{2\pi }\phi \left(z\right)dz=0$$

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Nov/18

$$t=tan\frac{\theta }{2}dt={sec}^2\frac{\theta }{2}\times \frac{1}{2}d\theta$$ $$d\theta =\frac{2dt}{1+t^2}$$ $$\int \frac{2dt}{(1+t^2)(a+b\times \frac{2t}{1+t^2}+c\times \frac{1-t^2}{1+t^2})}$$ $$\int \frac{2dt}{(a+{at}^2+2bt+c-{ct}^2)}$$ $$2\int \frac{dt}{t^2(a-c)+t\left(2b\right)+a+c}$$ $$\frac{2}{a-c}\int \frac{dt}{t^2+2.t.\frac{b}{a-c}+\frac{a+c}{a-c}}$$ $$\frac{2}{a-c}\int \frac{dt}{{(t+\frac{b}{a-c})}^2+\frac{a+c}{a-c}-\frac{b^2}{{(a-c)}^2}}$$ $$\frac{2}{a-c}\int \frac{dt}{{(t+\frac{b}{a-c})}^2+\frac{a^2-c^2-b^2}{{(a-c)}^2}}$$ $$\frac{2}{a-c}\times \frac{1}{\sqrt{\frac{a^2-b^2-c^2}{{(a-c)}^2}}}\times {tan}^{-1}\left(\frac{t+\frac{b}{a-c}}{\sqrt{\frac{a^2-b^2-c2}{{(a-c)}^2}}}\right)$$ $$=\frac{2}{\sqrt{a^2-b^2-c^2}}{tan}^{-1}\left(\frac{t+\frac{b}{a-c}}{\sqrt{\frac{a^2-b^2-c^2}{{(a-c)}^2}}}\right)$$ $$when\theta \to 0t\to 0$$ $$\theta \to 2\pi t\to 0$$ $$thusvalueofinyregationzero$$ $$plscheckwhereiamwrong..$$ $$since{a}^2>(b^2+c^2)$$ $$$$

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