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Question Number 35226 by abdo mathsup 649 cc last updated on 16/May/18

$$1)calculatef\left(a\right)={\int }_0^{\pi }\frac{dx}{a{sin}^{2}x+{cos}^{2}x}$$$$witha>0$$$$2)findthevalueofg\left(a\right)={\int }_0^{\pi }\frac{{sin}^{2}x}{{(a{sin}^{2}x+{cos}^{2}x)}^2}dx$$

Commented by prof Abdo imad last updated on 19/May/18

$$1)wehavef\left(a\right)={\int }_0^{\pi }\frac{dx}{a{sin}^{2}x+{cos}^{2}x}$$$$={\int }_0^{\pi }\frac{dx}{a\frac{1-cos\left(2x\right)}{2}+\frac{1+cos\left(2x\right)}{2}}$$$$=2{\int }_0^{\pi }\frac{dx}{a-acos\left(2x\right)+1+cos\left(2x\right)}$$$$=2{\int }_0^{\pi }\frac{dx}{a+1+(1-a)cos\left(2x\right)}$$$${=}_{2x=t}2{\int }_0^{2\pi }\frac{1}{1+a+(1-a)cost}\frac{dt}{2}$$$$={\int }_0^{2\pi }\frac{dt}{1+a+(1-a)cost}changement$$$$e^{it}=zgive$$$$I={\int }_{\mid z\mid =1}\frac{1}{1+a+(1-a)\frac{z+z^{-1}}{2}}\frac{dz}{iz}$$$$I={\int }_{\mid z\mid =1}\frac{2dz}{iz(2+2a+(1-a)(z+z^{-1})}$$$$={\int }_{\mid z\mid =1}\frac{-2idz}{2(1+a)z+(1-a)z^2+1-a}$$$$letconsider\phi \left(z\right)=\frac{-2i}{{(1-a)}^{}{z}^2+2(1+a)z+1-a}$$$$polesof\phi ?$$

Commented by prof Abdo imad last updated on 19/May/18

$${\mathrm{\Delta }}^{{}^{\prime }}={(1+a)}^2-{(1-a)}^2=1+2a+a^2-1+2a-a^2$$$$=4a\Rightarrow z_1=\frac{-1-a+2\sqrt{a}}{1-a}=\frac{a-2\sqrt{a}+1}{a-1}$$$$=\frac{{(\sqrt{a}-1)}^2}{(\sqrt{a}-1)(\sqrt{a}+1)}=\frac{\sqrt{a}-1}{\sqrt{a}+1}$$$$z_2=\frac{-1-a-2\sqrt{a}}{1-a}=\frac{{(\sqrt{a}+1)}^2}{a-1}=\frac{{(\sqrt{a}+1)}^2}{(\sqrt{a}-1)(\sqrt{a}+1)}$$$$=\frac{\sqrt{a}+1}{\sqrt{a}-1}$$$$\mid z_1\mid -1=\frac{\sqrt{a^{}}-1}{\sqrt{a}+1}-1=\frac{\sqrt{a}-1-\sqrt{a}-1}{\sqrt{a}+1}$$$$=\frac{-2}{\sqrt{a}+1}<0\Rightarrow \mid z_1\mid <1$$$$\mid z_2\mid -1=\frac{1}{\mid z_1\mid }-1=\frac{1-\mid z_1\mid }{\mid z_1\mid }>0\Rightarrow \mid z_2\mid >1(to$$$$elominatefromrrsidus)$$$${\int }_{\mid z\mid =1}\phi \left(z\right)dz=2i\pi Res(\phi ,z_1)$$$$Res(\phi ,z_1)={lim}_{z\to z_1}(z-z_1)\phi \left(z\right)$$$$but\phi \left(z\right)=\frac{-2i}{(1-a)(z-z_1)(z-z_2)}$$$$Res(\phi ,z_1)=\frac{-2i}{(1-a)(z_1-z_2)}=\frac{-2i}{(1-a)(z_1-\frac{1}{z_1})}$$$$=\frac{-2i{z}_1}{(1-a)(z_1^2-1)}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \frac{-2{iz}_1}{(1-a)(z_1^2-1)}$$$$=\frac{4\pi z_1}{(1-a)(z_1^2-1)}$$

Commented by abdo mathsup 649 cc last updated on 20/May/18

$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=\frac{4\pi }{1-a}\frac{\frac{\sqrt{a}-1}{\sqrt{a}+1}}{\frac{{(\sqrt{a}-1)}^2}{{(\sqrt{a}+1)}^2}-1}$$$$=\frac{4\pi }{1-a}\frac{\sqrt{a}-1}{\sqrt{a}+1}\frac{{(\sqrt{a}+1)}^2}{{(\sqrt{a}-1)}^2-{(\sqrt{a}+1)}^2}$$$$=\frac{4\pi (a-1)}{(1-a)(a-2\sqrt{a}+1-a-2\sqrt{a}-1)}$$$$=\frac{4\pi }{4\sqrt{a}}=\frac{\pi }{\sqrt{a}}so$$$$f\left(a\right)=\frac{\pi }{\sqrt{a}}.$$

Commented by abdo mathsup 649 cc last updated on 20/May/18

$$2)wehave{f}^{{}^{\prime }}\left(a\right)={\int }_0^{\pi }\frac{\partial }{\partial a}\left\{\frac{1}{{asin}^{2}x+{cos}^{2}x}\right\}dx$$$$=-{\int }_0^{\pi }\frac{{sin}^{2}x}{{(a{sin}^{2}x+{cos}^{2}x)}^2}dx=-g\left(a\right)\Rightarrow$$$$g\left(a\right)=-f^{{}^{\prime }}\left(a\right)butf\left(a\right)=\frac{\pi }{\sqrt{a}}\Rightarrow f^{{}^{\prime }}\left(a\right)=-\pi \frac{{\left(\sqrt{a}\right)}^{{}^{\prime }}}{a}$$$$=-\pi \frac{1}{2a\sqrt{a}}\Rightarrow g\left(a\right)=\frac{-\pi }{2a\sqrt{a}}.$$

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