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Question Number 64850 by mathmax by abdo last updated on 22/Jul/19

$$let{A}_{\lambda }={\int }_0^{\pi }\frac{dx}{\lambda +cosx+sinx}(\lambda \in R)$$$$1)findaexplicitformof{A}_{\lambda }$$$$2)findalso{B}_{\lambda }={\int }_0^{\pi }\frac{dx}{{(\lambda +cosx+sinx)}^2}$$$$3)calculate{\int }_0^{\pi }\frac{dx}{2+cosx+sinx}and{\int }_0^{\pi }\frac{dx}{{(3+cosx+sinx)}^2}$$

Commented by mathmax by abdo last updated on 23/Jul/19

$$1)A_{\lambda }={\int }_0^{\pi }\frac{dx}{\lambda +cosx+sinx}changementtan\left(\frac{x}{2}\right)=tgive$$$$A_{\lambda }={\int }_0^{\mathrm{\infty }}\frac{1}{\lambda +\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2}}\frac{2dt}{1+t^2}={\int }_0^{\mathrm{\infty }}\frac{2dt}{\lambda +\lambda t^2+1-t^2+2t}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{dt}{(\lambda -1)t^2+2t+1+\lambda }letdecomposeF\left(t\right)=\frac{1}{(\lambda -1)t^2+2t+1+\lambda }$$$${\mathrm{\Delta }}^{{}^{\prime }}=1-(\lambda -1)(\lambda +1)=1-({\lambda }^2-1)=2-{\lambda }^2$$$$case12-{\lambda }^2>0\Rightarrow \mid \lambda \mid <\sqrt{2}\Rightarrow t_1=\frac{-1+\sqrt{2-{\lambda }^2}}{\lambda -1}(\lambda \ne 1)$$$$t_2=\frac{-1-\sqrt{2-{\lambda }^2}}{\lambda -1}\Rightarrow F\left(t\right)=\frac{1}{(\lambda -1)(t-t_1)(t-t_2)}\Rightarrow$$$$A_{\lambda }=\frac{2}{\lambda -1}{\int }_0^{\mathrm{\infty }}\frac{1}{\frac{2\sqrt{2-{\lambda }^2}}{\lambda -1}}(\frac{1}{t-t_1}-\frac{1}{t-t_2})dt=\frac{1}{\sqrt{2-{\lambda }^2}}{\int }_0^{\mathrm{\infty }}(\frac{1}{t-t_1}-\frac{1}{t-t_2})dt$$$$=\frac{1}{2\sqrt{2-{\lambda }^2}}{[ln\mid \frac{t-t_1}{t-t_2}\mid ]}_0^{+\mathrm{\infty }}=\frac{1}{2\sqrt{2-{\lambda }^2}}ln\mid \frac{t_2}{t_1}\mid$$$$A_{\lambda }=\frac{1}{2\sqrt{2-{\lambda }^2}}ln\mid \frac{1+\sqrt{2-{\lambda }^2}}{1-\sqrt{2-{\lambda }^2}}\mid$$

Commented by mathmax by abdo last updated on 23/Jul/19

$$case22-{\lambda }^2<0\Rightarrow \mid \lambda \mid >\sqrt{2}\Rightarrow {\mathrm{\Delta }}^{{}^{\prime }}<0and$$$$(\lambda -1)t^2+2t+1+\lambda =(\lambda -1)\{t^2+\frac{2}{\lambda -1}t+\frac{1+\lambda }{\lambda -1}\}$$$$=(\lambda -1)\{t^2+\frac{2t}{\lambda -1}+\frac{1}{{(\lambda -1)}^2}+\frac{1+\lambda }{\lambda -1}-\frac{1}{{(\lambda -1)}^2}\}$$$$=(\lambda -1)\{{(t+\frac{1}{\lambda -1})}^2+\frac{{\lambda }^2-2}{{(\lambda -1)}^2}\}weusethechangement$$$$t+\frac{1}{\lambda -1}=\frac{\sqrt{{\lambda }^2-2}}{\mid \lambda -1\mid }u\Rightarrow A_{\lambda }=\frac{2}{\lambda -1}{\int }_0^{\mathrm{\infty }}\frac{dt}{{(t+\frac{1}{\lambda -1})}^2+\frac{{\lambda }^2-2}{{(\lambda -1)}^2}}$$$$=\frac{2}{\lambda -1}{\int }_{\frac{s\left(\lambda \right)}{\sqrt{{\lambda }^2-2}}}^{+\mathrm{\infty }}\frac{1}{\frac{{\lambda }^2-2}{{(\lambda -1)}^2}(1+u^2)}\frac{\sqrt{{\lambda }^2-2}}{\mid \lambda -1\mid }du$$$$=\frac{2{(\lambda -1)}^2}{(\lambda -1)\mid \lambda -1\mid }\sqrt{{\lambda }^2-2}{\left[arctanu\right]}_{\frac{s\left(\lambda \right)}{\sqrt{{\lambda }^2-2}}}^{+\mathrm{\infty }}$$$$A_{\lambda }=2s(\lambda -1)\sqrt{{\lambda }^2-2}(\frac{\pi }{2}-arctan\left(\frac{s\left(\lambda \right)}{\sqrt{{\lambda }^2-2}}\right))$$$$withs\left(x\right)=1ifx>0ands\left(x\right)=-1ifx<0$$

Commented by mathmax by abdo last updated on 23/Jul/19

$$2)wehave\frac{{dA}_{\lambda }}{d\lambda }=-{\int }_0^{\pi }\frac{dx}{{(\lambda +cosx+sinx)}^2}=-B_{\lambda }\Rightarrow$$$$B_{\lambda }=-A_{\lambda }^{{}^{\prime }}resttocalculate{A}_{\lambda }^{{}^{\prime }}....$$

Commented by mathmax by abdo last updated on 23/Jul/19

$$3)letcalculate{\int }_0^{\pi }\frac{dx}{2+cosx+sinx}here\lambda =2and$$$$2-{\lambda }^2=-2<0\Rightarrow {\int }_0^{\pi }\frac{dx}{2+cosx+sinx}=A_2$$$$=2s\left(1\right)\sqrt{2^2-2}\{\frac{\pi }{2}-arctan\left(\frac{s\left(2\right)}{\sqrt{2^2-2}}\right)\}$$$$=2\sqrt{2}\{\frac{\pi }{2}-arctan\left(\frac{1}{\sqrt{2}}\right)\}=\pi \sqrt{2}-2\sqrt{2}(\frac{\pi }{2}-arctan\left(\sqrt{2}\right))$$$$=\pi \sqrt{2}-\pi \sqrt{2}+2\sqrt{2}arctan\left(\sqrt{2}\right)\Rightarrow$$$${\int }_0^{\pi }\frac{dx}{2+cosx+sinx}=2\sqrt{2}arctan\left(\sqrt{2}\right).$$

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