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Question Number 33169 by abdo imad last updated on 11/Apr/18

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

Commented by abdo imad last updated on 12/Apr/18

$$letputI={\int }_0^{\pi }\frac{dx}{1+2{sin}^{2}x}={\int }_0^{\pi }\frac{dx}{1+2(1-{cos}^{2}x)}$$$$={\int }_0^{\pi }\frac{dx}{3-2{cos}^{2}x}={\int }_0^{\frac{\pi }{2}}\frac{dx}{3-2{cos}^{2}x}+{\int }_{\frac{\pi }{2}}^{\pi }\frac{dx}{3-2{cos}^{2}x}$$$$=I_1+I_{2}butweknewthat1+{tan}^{2}t=\frac{1}{{cos}^{2}t}\Rightarrow$$$${cos}^{2}x=\frac{1}{1+{tan}^{2}x}\Rightarrow I_1={\int }_0^{\frac{\pi }{2}}\frac{dx}{3-\frac{2}{1+{tan}^{2}x}}$$$$={\int }_0^{\frac{\pi }{2}}\frac{1+{tan}^{2}x}{1+3{tan}^{2}x}dx={}_{tanx=t}{\int }_0^{+\mathrm{\infty }}\frac{1+t^2}{1+3t^2}\frac{dt}{1+t^2}$$$$={\int }_0^{\mathrm{\infty }}\frac{dt}{1+3t^2}{=}_{t\sqrt{3}=u}{\int }_0^{\mathrm{\infty }}\frac{1}{1+u^2}\frac{du}{\sqrt{3}}=\frac{1}{\sqrt{3}}{\left[arctanu\right]}_0^{+\mathrm{\infty }}$$$$I_1=\frac{\pi }{2\sqrt{3}}letfind{I}_2?$$$$I_2={\int }_{\frac{\pi }{2}}^{\pi }\frac{dx}{3-2{cos}^{2}x}{=}_{x=\pi -t}{\int }_{\frac{\pi }{2}}^0\frac{-dt}{3-2{cos}^{2}t}={\int }_0^{\frac{\pi }{2}}\frac{dt}{3-2{cos}^{2}t}$$$$=I_1=\frac{\pi }{2\sqrt{3}}\Rightarrow I=\frac{\pi }{2\sqrt{3}}+\frac{\pi }{2\sqrt{3}}=\frac{\pi }{\sqrt{3}}.$$$$★I=\frac{\pi }{\sqrt{3}}★$$

Answered by sma3l2996 last updated on 11/Apr/18

$$I={\int }_0^{\pi }\frac{dx}{1+2{sin}^{2}x}$$$$t=tanx\Rightarrow dx=\frac{dt}{1+t^2}$$$${sin}^{2}x=\frac{t^2}{1+t^2}$$$$I={\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^2+2t^2}={\int }_0^{\mathrm{\infty }}\frac{dt}{3t^2+1}$$$$u=\sqrt{3}x\Rightarrow dx=\frac{\sqrt{3}}{3}du$$$$I=\frac{\sqrt{3}}{3}{\int }_0^{\mathrm{\infty }}\frac{du}{u^2+1}=\frac{\sqrt{3}}{3}{\left[{tan}^{-1}\left(u\right)\right]}_0^{\mathrm{\infty }}$$$$I=\frac{\sqrt{3}}{6}\pi$$

Commented by abdo imad last updated on 12/Apr/18

$$youranswerisnotcorectsirsma3l...$$

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