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Question Number 50412 by Abdo msup. last updated on 16/Dec/18

$$1)calculate{U}_n={\int }_0^{\pi }\frac{dx}{1+{cos}^2\left(nx\right)}withnfromN$$$$2)fcontinuefrom[0,\pi ]toRfind$$$${lim}_{n\to +\mathrm{\infty }}{\int }_0^{\pi }\frac{f\left(x\right)}{1+{cos}^2\left(nx\right)}dx$$

Commented by Abdo msup. last updated on 25/Dec/18

$$1)changementnx=tgive$$$$U_n=\frac{1}{n}{\int }_0^{n\pi }\frac{1}{1+{cos}^{2}t}dt\Rightarrow {nU}_n=\sum _{k=0}^{n-1}{\int }_{k\pi }^{(k+1)\pi }\frac{dt}{1+{cos}^{2}t}$$$$but{\int }_{k\pi }^{(k+1)\pi }\frac{dt}{1+{cos}^{2}t}{=}_{t=k\pi +x}{\int }_0^{\pi }\frac{dx}{1+{cos}^{2}x}$$$$={\int }_0^{\pi }\frac{dx}{1+\frac{1+cos\left(2x\right)}{2}}={\int }_0^{\pi }\frac{2dx}{3+cos\left(2x\right)}$$$${=}_{2x=u}{\int }_0^{2\pi }\frac{du}{3+cosu}du\Rightarrow {nU}_n=n{\int }_0^{2\pi }\frac{du}{3+cosu}$$$$\Rightarrow U_n={\int }_0^{2\pi }\frac{du}{3+cos\left(u\right)}changement{e}^{iu}=zgive$$$$U_n={\int }_{\mid z\mid =1}\frac{1}{3+\frac{z+z^{-1}}{2}}\frac{dz}{iz}$$$$={\int }_{\mid z\mid =1}\frac{2dz}{iz(6+z+z^{-1})}={\int }_{\mid z\mid =1}\frac{-2idz}{6z+z^2+1}$$$$let\phi \left(z\right)=\frac{-2i}{z^2+6z+1}polesof\phi ?$$$${\mathrm{\Delta }}^{{}^{\prime }}=3^2-1=8\Rightarrow z_1=-3+2\sqrt{2}and{z}_2=-3-2\sqrt{2}$$$$\mid z_1\mid -1=3-2\sqrt{2}-1=2-2\sqrt{2}<0\Rightarrow \mid z_1\mid <1$$$$\mid z_2\mid -1=3+2\sqrt{2}-1=2+2\sqrt{2}\left(z_{2}isoutofcircle\right)$$$${\int }_{\mid z\mid =1}\phi \left(z\right)dz=2i\pi Res(\phi ,z_1)$$$$Res(\phi ,z_1)=\frac{-2i}{z_1-z_2}=\frac{-2i}{4\sqrt{2}}=\frac{-i}{2\sqrt{2}}\Rightarrow$$$${\int }_{\mid z\mid =1}\phi \left(z\right)dz=2i\pi \left(\frac{-i}{2\sqrt{2}}\right)=\frac{\pi }{\sqrt{2}}$$$$\Rightarrow \mathrm{\forall }n\in N{U}_n=\frac{\pi }{\sqrt{2}}.$$

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