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Question Number 200319 by sonukgindia last updated on 17/Nov/23

Answered by Sutrisno last updated on 17/Nov/23

$$={\int }_0^{\pi }\frac{\frac{1}{{cos}^{2}x}}{\frac{1}{{cos}^{2}x}+\frac{{sin}^{2}x}{{cos}^{2}x}}dx$$$$={\int }_0^{\pi }\frac{{sec}^{2}x}{{sec}^{2}x+{tan}^{2}x}dx$$$$={\int }_0^{\pi }\frac{{sec}^{2}x}{1+2{tan}^{2}x}dx$$$$misal:tan\theta =\sqrt{2}tanx\to \frac{{sec}^2\theta }{\sqrt{2}{sec}^{2}x}d\theta =dx$$$$=\int \frac{{sec}^{2}x}{1+2{\left(\frac{tan\theta }{\sqrt{2}}\right)}^2}.\frac{{sec}^2\theta }{\sqrt{2}{sec}^{2}x}d\theta$$$$=\frac{1}{\sqrt{2}}\int \frac{{sec}^2\theta }{1+{tan}^2\theta }d\theta$$$$=\frac{1}{\sqrt{2}}\int d\theta$$$$=\frac{1}{\sqrt{2}}\theta$$$$=\frac{1}{\sqrt{2}}arctan\left(\sqrt{2}tanx\right){\mid }_0^{\pi }$$$$=\frac{1}{\sqrt{2}}[arctan\left(\sqrt{2}tan\pi \right)-arctan\left(\sqrt{2}tan0\right)]$$$$=0$$$$$$

Commented by mr W last updated on 17/Nov/23

$$\frac{1}{2}<\frac{1}{1+{\sin }^{2}x}<1$$$$\Rightarrow \frac{\pi }{2}<{\int }_0^{\pi }\frac{dx}{1+{\sin }^{2}x}<\pi$$$$thatmeans{\int }_0^{\pi }\frac{dx}{1+{\sin }^{2}x}canneverbe0.$$

Answered by MM42 last updated on 17/Nov/23

$$I={\int }_0^{\pi }\frac{1}{1+{sin}^{2}x}dx=2{\int }_0^{\frac{\pi }{2}}\frac{1}{1+{sin}^{2}x}dx$$$$=2{\int }_0^{\frac{\pi }{2}}\frac{1+{cot}^{2}x}{2+{cot}^{2}x}dx;cotx=u$$$${\Rightarrow I=2{\int }_0^{\mathrm{\infty }}\frac{du}{2+u^2}du=\sqrt{2}{tan}^{-1}\left(\frac{u}{\sqrt{2}}\right)]}_0^{\mathrm{\infty }}$$$$=\frac{\sqrt{2}}{2}\pi ✓$$$$$$

Answered by Mathspace last updated on 18/Nov/23

$$$$$$I={\int }_0^{\pi }\frac{dx}{1+{sin}^{2}x}={\int }_0^{\pi }\frac{dx}{1+\frac{1-cos\left(2x\right)}{2}}$$$$=2{\int }_0^{\pi }\frac{dx}{3-cos\left(2x\right)}(2x=t)$$$$={\int }_0^{2\pi }\frac{dt}{3-cost}={\int }_0^{2\pi }\frac{dt}{3-\frac{e^{it}+e^{-it}}{2}}$$$$={\int }_0^{2\pi }\frac{2dt}{6-e^{it}-e^{-it}}(e^{it}=z)$$$$={\int }_{\mid z\mid =1}\frac{2dz}{iz(6-z-z^{-1})}$$$$={\int }_{\mid z\mid =1}\frac{-2idz}{6z-z^2-1}$$$$={\int }_{\mid z\mid =1}\frac{2idz}{z^2-6z+1}$$$$f\left(z\right)=\frac{2i}{z^2-6z+1}lespoles?$$$${\mathrm{\Delta }}^{{}^{\prime }}=3^2-1=8\Rightarrow$$$$z_1=3+\sqrt{8}=3+2\sqrt{2}$$$$z_2=3-\sqrt{8}=3-2\sqrt{2}$$$$\mid z_1\mid -1=3+2\sqrt{2}-1=2+2\sqrt{2}>0$$$$\mid z_2\mid -1=3-2\sqrt{2}-1=2-2\sqrt{2}$$$$<0\Rightarrow$$$${\int }_{\mid z\mid }f\left(z\right)dz=2i\pi Res(f,z_2)$$$$f\left(z\right)=\frac{2i}{(z-z_1)(z-z_2)}\Rightarrow$$$$Res(f,z_2)=\frac{2i}{z_2-z_1}=\frac{2i}{-4\sqrt{2}}=-\frac{i}{2\sqrt{2}}$$$${\int }_{\mid z\mid =1}f\left(z\right)dz=2i\pi \times \frac{-i}{2\sqrt{2}}$$$$=\frac{\pi }{\sqrt{2}}\Rightarrow ★I=\frac{\pi }{\sqrt{2}}★$$

Answered by mnjuly1970 last updated on 18/Nov/23

$$\mathrm{I}=2{\int }_0^{\frac{\pi }{2}}\frac{1+{tan}^2\left(x\right)}{1+2{tan}^2\left(x\right)}dx\stackrel{tanx=y}{=}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{dy}{1+2y^2}={\int }_0^{\mathrm{\infty }}\frac{dy}{{\left(\frac{\sqrt{2}}{2}\right)}^2+y^2}$$$$={\left[\sqrt{2}{tan}^{-1}\left(y\sqrt{2}\right)\right]}_0^{\mathrm{\infty }}=\frac{\pi \sqrt{2}}{2}$$

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