Question Number 74514 by mathmax by abdo last updated on 25/Nov/19
$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{\left({x}−{sin}\theta\right){d}\theta}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}\:{sin}\theta\:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Commented by mathmax by abdo last updated on 26/Nov/19
$${we}\:{have}\:{proved}\:{that}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dt}}{{x}^{\mathrm{2}} −\mathrm{2}{xsin}\left(\mathrm{2}{t}\right)+\mathrm{1}}\:=\frac{\pi}{\mathrm{1}−{x}^{\mathrm{2}} }\:{if}\:\mid{x}\mid<\mathrm{1}\:{and} \\ $$$$=\frac{\pi}{{x}^{\mathrm{2}} −\mathrm{1}}\:{if}\:\mid{x}\mid>\mathrm{1}\:\:{let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\pi} \:\frac{{dt}}{{x}^{\mathrm{2}} −\mathrm{2}{x}\:{sin}\left(\mathrm{2}{t}\right)+\mathrm{1}}\:\Rightarrow \\ $$$${f}\left({x}\right)=_{\mathrm{2}{t}\:=\theta} \:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{d}\theta}{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{2}{xsin}\theta\:+\mathrm{1}\right)}\:\Rightarrow\mathrm{2}{f}^{'} \left({x}\right)=−\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{2}{x}−\mathrm{2}{sin}\theta}{\left({x}^{\mathrm{2}} \left(\mathrm{2}{xsin}\theta\:+\mathrm{1}\right)^{\mathrm{2}} \right.}{d}\theta\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\left({x}−{sin}\theta\right)}{\left({x}^{\mathrm{2}} −\mathrm{2}{xsin}\theta\:+\mathrm{1}\right)^{\mathrm{2}} }{d}\theta\:=−{f}^{'} \left({x}\right) \\ $$$$\mid{x}\mid<\mathrm{1}\:\Rightarrow{f}^{'} \left({x}\right)=−\pi\frac{−\mathrm{2}{x}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{\mathrm{2}\pi{x}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\mid{x}\mid>\mathrm{1}\:\Rightarrow{f}^{'} \left({x}\right)=−\frac{\pi\:\left(\mathrm{2}{x}\right)}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{−\mathrm{2}\pi{x}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:{so}\:{the}\:{value}\:{of}\:{this}\:{integral} \\ $$$${is}\:{known}. \\ $$
Answered by mind is power last updated on 25/Nov/19
$${use}\:{provious}\:{Quation} \\ $$$${f}\left({x}\right)=\int_{\mathrm{0}} ^{\pi} \frac{{d}\theta}{{x}^{\mathrm{2}} −\mathrm{2}{sin}\left(\mathrm{2}\theta\right){x}+\mathrm{1}} \\ $$$${u}=\mathrm{2}\theta \\ $$$${f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{du}}{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{2}{sin}\left({u}\right){x}+\mathrm{1}\right)} \\ $$$$\Rightarrow\mathrm{2}{f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{du}}{{x}^{\mathrm{2}} −\mathrm{2}{sin}\left({u}\right){x}+\mathrm{1}}=\begin{cases}{\frac{\pi{x}}{{x}^{\mathrm{2}} −\mathrm{1}},\:\:{x}>\mathrm{1}}\\{\frac{\pi}{\mathrm{1}−{x}^{\mathrm{2}} },\:\:\:\mathrm{0}<{x}<\mathrm{1}}\end{cases} \\ $$$$\mathrm{2}{f}'\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{−\mathrm{2}{x}+\mathrm{2}{sin}\left({u}\right)}{\left({x}^{\mathrm{2}} −\mathrm{2}{xsin}\left({u}\right)+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow−{f}'\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{x}−{sin}\left({u}\right)}{\left({x}^{\mathrm{2}} −\mathrm{2}{xsin}\left({u}\right)+\mathrm{1}\right)^{\mathrm{2}} }{du} \\ $$$${f}'\left({x}\right)=\pi.\left(\frac{−{x}^{\mathrm{2}} −\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\right)\:\:\:\:\:\:\mathrm{0}\leqslant{x}<\mathrm{1} \\ $$$${f}'\left({x}\right)=\frac{\mathrm{2}{x}\pi}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} },{x}>\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 26/Nov/19
$${thank}\:{you}\:{sir}. \\ $$