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Question Number 31071 by abdo imad last updated on 02/Mar/18

$$find{\int }_0^{\pi }\frac{dx}{1+{sin}^{2}x}.$$

Commented by abdo imad last updated on 03/Mar/18

$$thanks.$$

Commented by abdo imad last updated on 02/Mar/18

$$letputI={\int }_0^{\pi }\frac{dx}{1+{sin}^{2}x}\Rightarrow I={\int }_0^{\pi }\frac{dx}{1+\frac{1-cos\left(2x\right)}{2}}$$$$={\int }_0^{\pi }\frac{2dx}{3-cos\left(2x\right)}{=}_{t=2x}{\int }_0^{2\pi }\frac{dt}{3-cost}thech.e^{it}=zgive$$$$I={\int }_{\mid z\mid =1}\frac{1}{3-\frac{z+z^{-1}}{2}}\frac{dz}{iz}={\int }_{\mid z\mid =1}\frac{2dz}{iz(6-(z+z^{-1}))}$$$$={\int }_{\mid z\mid =1}\frac{-2idz}{6z-z^2-1}={\int }_{\mid z\mid =1}\frac{2idz}{z^2-6z+1}letput$$$$\phi \left(z\right)=\frac{2i}{z^2-6z+1}.polesof\phi ?$$$$z^2-6z+1=0\Rightarrow {\mathrm{\Delta }}^{{}^{\prime }}={(-3)}^2-1=8\Rightarrow z_1=3+2\sqrt{2}$$$$z_2=3-2\sqrt{2}\mid z_1\mid -1=3+2\sqrt{2}-1=2+2\sqrt{2}>0(toeliminate$$$$fromresidus)\mid z_2\mid -1=2-2\sqrt{2}=2(1-\sqrt{2)}<0\Rightarrow$$$${\int }_{\mid z\mid =1}\phi \left(z\right)dz=2i\pi Res(\phi ,z_2)but$$$$\phi \left(z\right)=\frac{2i}{(z-z_1)(z-z_2)}\Rightarrow Res(\phi ,z_2)=\frac{2i}{z_2-z_1}=\frac{2i}{-4\sqrt{2}}=\frac{-i}{2\sqrt{2}}$$$${\int }_{\mid z\mid =1}\phi \left(z\right)dz=2i\pi .\frac{-i}{2\sqrt{2}}=\frac{\pi }{\sqrt{2}}\Rightarrow I=\frac{\pi }{\sqrt{2}}.$$$$$$$$$$

Commented by Joel578 last updated on 02/Mar/18

$$\mathrm{Nice}\mathrm{solution},\mathrm{Sir}$$

Answered by Joel578 last updated on 02/Mar/18

$$I={\int }_0^{\pi }\frac{{\mathrm{cosec}}^{2}x}{(1+{\sin }^{2}x){\mathrm{cosec}}^{2}x}dx$$$$={\int }_0^{\pi }\frac{{\mathrm{cosec}}^{2}x}{{\mathrm{cosec}}^{2}x+1}dx$$$$={\int }_0^{\pi }\frac{{\mathrm{cosec}}^{2}x}{{\cot }^{2}x+2}dx$$$$$$$$u=\cot x\to du=-{\mathrm{cosec}}^{2}xdx$$$$=-\int \frac{{\mathrm{cosec}}^{2}x}{u^2+2}.\frac{du}{{\mathrm{cosec}}^{2}x}=-\int \frac{du}{u^2+2}$$$$$$$$u=\sqrt{2}\tan \theta \to du=\sqrt{2}{\sec }^2\theta d\theta$$$$I=-\sqrt{2}\int \frac{{\sec }^2\theta }{2({\tan }^2\theta +1)}d\theta$$$$=-\frac{\sqrt{2}}{2}\int d\theta$$$$=-\frac{\sqrt{2}}{2}\theta +C=-\frac{\sqrt{2}}{2}{\tan }^{-1}\left(\frac{u}{\sqrt{2}}\right)+C$$$$=-\frac{\sqrt{2}}{2}{\tan }^{-1}\left(\frac{\cot x}{\sqrt{2}}\right)+C$$$$$$$$I={[-\frac{\sqrt{2}}{2}{\tan }^{-1}\left(\frac{\cot x}{\sqrt{2}}\right)]}_0^{\pi }$$$$=(-\frac{\sqrt{2}}{2}{\tan }^{-1}(-\mathrm{\infty }))-(-\frac{\sqrt{2}}{2}{\tan }^{-1}\left(\mathrm{\infty }\right))$$$$=\frac{\pi \sqrt{2}}{4}+\frac{\pi \sqrt{2}}{4}=\frac{\pi \sqrt{2}}{2}$$

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