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Question Number 38210 by prof Abdo imad last updated on 22/Jun/18

$$letf\left(a\right)={\int }_0^{\pi }\frac{d\theta }{a+{sin}^2\theta }\left(afromR\right)$$$$1)findf\left(a\right)$$$$2)calculateg\left(a\right)={\int }_0^{\pi }\frac{d\theta }{{(a+{sin}^2\theta )}^2}$$$$3)calculate{\int }_0^{\pi }\frac{d\theta }{1+{sin}^2\theta }and{\int }_0^{\pi }\frac{d\theta }{2+{sin}^2\theta }$$$$4)calculate{\int }_0^{\pi }\frac{d\theta }{{(3+{sin}^2\theta )}^2}.$$

Commented by math khazana by abdo last updated on 23/Jun/18

$$wehavef\left(a\right)={\int }_0^{\pi }\frac{d\theta }{a+\frac{1-cos\left(2\theta \right)}{2}}$$$$={\int }_0^{\pi }\frac{2d\theta }{2a+1-cos\left(2\theta \right)}{=}_{2\theta =t}{\int }_0^{2\pi }\frac{2}{2a+1-cost}\frac{dt}{2}$$$$={\int }_0^{2\pi }\frac{dt}{2a+1-cost}changement{e}^{it}=zgive$$$$f\left(a\right)={\int }_{\mid z\mid =1}\frac{1}{2a+1-\frac{z+z^{-1}}{2}}\frac{dz}{iz}$$$$={\int }_{\mid z\mid =1}\frac{-2idz}{z\{4a+2-z-z^{-1}\}}$$$$={\int }_{\mid z\mid =1}\frac{-2idz}{(4a+2)z-z^2-1}={\int }_{\mid z\mid =1}\frac{2idz}{z^2-(4a+2)z+1}$$$$let\phi \left(z\right)=\frac{2i}{z^2-(4a+2)z+1}.polesof\phi ?$$$${\mathrm{\Delta }}^{{}^{\prime }}={(2a+1)}^2-1=4a^2+4a=4a(a+1)$$$$case1a(a+1)>0\Rightarrow z_1=2a+1+2\sqrt{a^2+a}$$$$z_2=2a+1-2\sqrt{a^2+a}$$$$\mid z_2\mid -1=2a-2\sqrt{a^2+a}<0anditsclearthat$$$$\mid z_1\mid -1>0\left(toelominatefromresidus\right)$$$${\int }_{\mid z\mid =1}\phi \left(z\right)dz=2i\pi Res(\phi ,z_2)=\frac{2i}{z_2-z_1}2i\pi$$$$=\frac{-4\pi }{-4\sqrt{a^2+a}}=\frac{\pi }{\sqrt{a^2+a}}\Rightarrow$$$$f\left(a\right)=\frac{\pi }{\sqrt{a^2+a}}$$$$case2a(a+1)<0\iff -1<a<0\Rightarrow {\mathrm{\Delta }}^{,}=-4(-a(a+1))$$$$={\left(2i\sqrt{-a^2-a}\right)}^2$$$$z_1=2a+1+2i\sqrt{-a^2-a}$$$$z_2=2a+1-2i\sqrt{-a^2-a}$$$$\mid z_1\mid =\sqrt{{(2a+1)}^2+4(-a^2-a)}$$$$=\sqrt{4a^2+4a+1-4a^2-4a}=1$$$$\mid z_2\mid =1\Rightarrow$$$${\int }_{\mid z\mid =1}\phi \left(z\right)dz=2i\pi \{Res(\phi ,z_1)+Res(\phi ,z_2)\}$$$$Res(\phi ,z_1)=\frac{2i}{z_1-z_2}$$$$Res(\phi ,z_2)=\frac{2i}{z_2-z_1}\Rightarrow {\int }_{\mid z\mid =1}\phi \left(z\right)dz=0\Rightarrow f\left(a\right)=0$$

Commented by math khazana by abdo last updated on 23/Jun/18

$$2)wehave{f}^{{}^{\prime }}\left(a\right)={\int }_0^{\pi }\frac{\partial }{\partial a}\left(\frac{1}{a+{sin}^2\theta }\right)d\theta$$$$=-{\int }_0^{\pi }\frac{d\theta }{{(a+{sin}^2\theta )}^2}=-g\left(a\right)\Rightarrow$$$$g\left(a\right)=-f^{{}^{\prime }}\left(a\right)$$$$ifa(a+1)<0\Rightarrow g\left(a\right)=0$$$$ifa(a+1)>0f\left(a\right)=\frac{\pi }{\sqrt{a^2+a}}\Rightarrow$$$$f^{{}^{\prime }}\left(a\right)=\pi {\left\{{(a^2+a)}^{-\frac{1}{2}}\right\}}^{{}^{\prime }}=-\frac{\pi }{2}(2a+1){(a^2+a)}^{-\frac{3}{2}}$$$$=\frac{-\pi (2a+1)}{2(a^2+a)\sqrt{a^2+a}}\Rightarrow$$$$g\left(a\right)=\frac{(2a+1)\pi }{2(a^2+a)\sqrt{a^2+a}}.$$$$$$

Commented by math khazana by abdo last updated on 23/Jun/18

$$3)wehaveprovedthatf\left(a\right)=\frac{\pi }{\sqrt{a^2+a}}ifa(a+1)>0$$$$so{\int }_0^{\pi }\frac{d\theta }{1+{sin}^2\theta }=f\left(1\right)=\frac{\pi }{\sqrt{2}}$$$${\int }_0^{\pi }\frac{d\theta }{2+{sin}^2\theta }=f\left(2\right)=\frac{\pi }{\sqrt{6}}$$$$$$$$\frac{}{}$$

Commented by math khazana by abdo last updated on 23/Jun/18

$$4){\int }_0^{\pi }\frac{d\theta }{{(3+{sin}^2\theta )}^2}=g\left(3\right)=\frac{7\pi }{24\sqrt{12}}=\frac{7\pi }{48\sqrt{3}}.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jun/18

$${\int }_0^{\mathrm{\Pi }}\frac{d\theta }{a+{sin}^2\theta }$$$$\int \frac{{sec}^2\theta }{{asec}^2\theta +{tan}^2\theta }d\theta$$$$t=tan\theta dt={sec}^2\theta d\theta$$$$\int \frac{dt}{a+{at}^2+t^2}$$$$\int \frac{dt}{a+t^2(1+a)}$$$$\frac{1}{1+a}\int \frac{dt}{\frac{a}{1+a}+t^2}$$$$\frac{1}{1+a}\times \frac{1}{\frac{\sqrt{a}}{\sqrt{1+a}}}{tan}^{-1}\left(\frac{t}{\frac{\sqrt{a}}{\sqrt{1+a}}}\right)$$$$=2\mid \frac{1}{\sqrt{a}\times \sqrt{1+a}}\times {tan}^{-1}\left(\frac{tan\theta }{\frac{\sqrt{a}}{\sqrt{1+a}}}\right){\mid }_0^{\frac{\mathrm{\Pi }}{2}}$$$$=2\times \frac{1}{\sqrt{a}\times \sqrt{1+a}}\times \frac{\mathrm{\Pi }}{2}$$$$=\frac{\mathrm{\Pi }}{\sqrt{a+a^2}}\mathcal{A}ns$$$$$$$$$$$${\int }_0^{\mathrm{\Pi }}f\left(\theta \right)d\theta =2{\int }_0^{\frac{\mathrm{\Pi }}{2}}f\left(\theta \right)d\theta whenf(\mathrm{\Pi }-\theta )=f\left(\theta \right)$$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jun/18

$$g\left(a\right)={\int }_0^{\mathrm{\Pi }}\frac{d\theta }{{(a+{sin}^2\theta )}^2}$$$$f\left(a\right)={\int }_0^{\mathrm{\Pi }}\frac{d\theta }{a+{sin}^2\theta }$$$$\frac{df\left(a\right)}{da}={\int }_0^{\mathrm{\Pi }}\frac{\partial }{\partial a}\frac{1}{(a+{sin}^2\theta )}d\theta$$$$={\int }_0^{\mathrm{\Pi }}\frac{-1}{{(a+{sin}^2\theta )}^2}d\theta$$$$so\frac{df\left(a\right)}{da}=-g\left(a\right)$$$$g\left(a\right)=-\frac{d}{da}\left(\frac{\mathrm{\Pi }}{\sqrt{a+a^2}}\right)$$$$g\left(a\right)=-\mathrm{\Pi }\times \frac{-1}{2{(a+a^2)}^{\frac{3}{2}}}\times (1+2a)$$$$g\left(a\right)=\frac{\mathrm{\Pi }(1+2a)}{2{(a+a^2)}^{\frac{3}{2}}}ANS$$$$ihaveproved{\int }_0^{\mathrm{\Pi }}\frac{d\theta }{a+{sin}^2\theta }=\frac{\mathrm{\Pi }}{\sqrt{a+a^2}}$$$$$$

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