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Question Number 46851 by maxmathsup by imad last updated on 01/Nov/18

$$letf\left(x\right)={\int }_0^{2\pi }\frac{sint}{x+sint}dtwithx>1$$$$1)calculatef\left(x\right)$$$$2)calculate{\int }_0^{2\pi }\frac{sint}{{(x+sint)}^2}dt$$$$3)findthevalueof{\int }_0^{2\pi }\frac{sint}{2+sint}dtand{\int }_0^{2\pi }\frac{sint}{{(2+sint)}^2}dt$$

Commented by maxmathsup by imad last updated on 02/Nov/18

$$1)wehavef\left(x\right)={\int }_0^{2\pi }\frac{x+sint-x}{x+sint}dt=2\pi -x{\int }_0^{2\pi }\frac{dt}{x+sint}$$$$changement{e}^{it}=zgive{\int }_0^{2\pi }\frac{dt}{x+sint}={\int }_{\mid z\mid =1}\frac{1}{x+\frac{z-z^{-1}}{2i}}\frac{dz}{iz}$$$$={\int }_{\mid z\mid =1}\frac{2idz}{iz(2ix+z-z^{-1})}={\int }_{\mid z\mid =1}\frac{2idz}{-2xz+{iz}^2-i}={\int }_{\mid z\mid =1}\frac{2dz}{2ixz+z^2-1}$$$$={\int }_{\mid z\mid =1}\frac{2dz}{z^2+2ixz-1}let\phi \left(z\right)=\frac{2}{z^2+2ixz-1}polesof\phi ?$$$${\mathrm{\Delta }}^{{}^{\prime }}={\left(ix\right)}^2+1=1-x^2={\left(i\sqrt{x^2-1}\right)}^2\Rightarrow z_1=-ix+i\sqrt{x^2-1}$$$$z_2=-ix-i\sqrt{x^2-1}$$$$\mid z_1\mid -1=\mid -x+\sqrt{x^2-1}\mid -1=\sqrt{x^2-1}-x-1=\sqrt{x^2-1}-(x+1)$$$$and{x}^2-1-{(x+1)}^2=x^2-1-x^2-2x-1=-2x-2<0\Rightarrow \mid z_1\mid <1$$$$\mid z_2\mid -1=\mid x+\sqrt{x^2-1}\mid -1=x-1+\sqrt{x^2-1}>0becausex>1\Rightarrow z_{2}isoutofcircle$$$$so{\int }_{\mid z\mid =1}\phi \left(z\right)dz=2i\pi Res(\phi ,z_1)but\phi \left(z\right)=\frac{2}{(z-z_1)(z-z_2)}\Rightarrow$$$$Res(\phi ,z_1)={lim}_{z\to z_1}(z-z_1)\phi \left(z\right)=\frac{2}{z_1-z_2}=\frac{2}{2i\sqrt{x^2-1}}\Rightarrow$$$${\int }_{\mid z\mid =1}\phi \left(z\right)dz=2i\pi \frac{1}{i\sqrt{x^2-1}}=\frac{2\pi }{\sqrt{x^2-1}}\Rightarrow ★f\left(x\right)=\frac{2\pi }{\sqrt{x^2-1}}★withx>1$$

Commented by maxmathsup by imad last updated on 02/Nov/18

$$2)wehave{f}^{{}^{\prime }}\left(x\right)={\int }_0^{2\pi }\frac{\partial }{\partial x}\left\{\frac{sint}{x+sint}\right\}dt=-{\int }_0^{2\pi }\frac{sint}{{(x+sint)}^2}dt\Rightarrow$$$${\int }_0^{2\pi }\frac{sint}{{(x+sint)}^2}dt=-f^{{}^{\prime }}\left(x\right)butf\left(x\right)=\frac{2\pi }{\sqrt{x^2-1}}=2\pi {(x^2-1)}^{-\frac{1}{2}}\Rightarrow$$$$f^{{}^{\prime }}\left(x\right)=2\pi (-\frac{1}{2})\left(2x\right){(x^2-1)}^{-\frac{3}{2}}=\frac{-2\pi x}{{(x^2-1)}^{\frac{3}{2}}}=\frac{-2\pi x}{(x^2-1)\sqrt{x^2-1}}\Rightarrow$$$${\int }_0^{2\pi }\frac{sint}{{(x+sint)}^2}dt=\frac{2\pi x}{(x^2-1)\sqrt{x^2-1}}withx>1.$$

Commented by maxmathsup by imad last updated on 02/Nov/18

$$3)wehave{\int }_0^{2\pi }\frac{sint}{2+sint}dt=f\left(2\right)=\frac{2\pi }{\sqrt{2^2-1}}=\frac{2\pi }{\sqrt{3}}$$$${\int }_0^{2\pi }\frac{sint}{{(2+sint)}^2}dt=f^{{}^{\prime }}\left(2\right)=\frac{4\pi }{(2^2-1)\sqrt{2^2-1}}=\frac{4\pi }{3\sqrt{3}}.$$

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