Question Number 201293 by sonukgindia last updated on 03/Dec/23
Answered by aleks041103 last updated on 03/Dec/23
$${I}=\int_{\mathrm{2}} ^{\:\infty} \frac{\mathrm{8}{arcsec}\left({x}/\mathrm{2}\right){dx}}{{x}^{\mathrm{3}} −\mathrm{4}{x}}= \\ $$$$=\int_{\mathrm{1}} ^{\:\infty} \frac{\mathrm{8}{arcsec}\left(\left(\mathrm{2}{x}\right)/\mathrm{2}\right)}{\left(\mathrm{2}{x}\right)^{\mathrm{3}} −\mathrm{4}\left(\mathrm{2}{x}\right)}{d}\left(\mathrm{2}{x}\right)=\mathrm{2}\int_{\mathrm{1}} ^{\:\infty} \frac{{arcsec}\left({x}\right)}{{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{dx} \\ $$$${x}={sec}\left({t}\right)\Rightarrow{dx}={sec}\left({t}\right){tan}\left({t}\right){dt} \\ $$$$\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \frac{{t}\:{sec}\left({t}\right)\:{tan}\left({t}\right)\:{dt}}{{sec}\left({t}\right)\left({sec}^{\mathrm{2}} \left({t}\right)−\mathrm{1}\right)}= \\ $$$$=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \frac{{t}\:{cos}\left({t}\right)}{{sin}\left({t}\right)}{dt}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {td}\left({ln}\left({sin}\left({t}\right)\right)\right)= \\ $$$$=\left[{tln}\left({sin}\left({t}\right)\right)\right]_{\mathrm{0}} ^{\pi/\mathrm{2}} −\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({sin}\left({t}\right)\right){dt} \\ $$$$=−\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({sin}\left({t}\right)\right){dt} \\ $$$${J}=\int_{\mathrm{0}} ^{\:\:\pi/\mathrm{2}} {ln}\left({sin}\left({t}\right)\right){dt}=\int_{\pi/\mathrm{2}} ^{\:\:\mathrm{0}} {ln}\left({sin}\left(\frac{\pi}{\mathrm{2}}−{s}\right)\right){d}\left(\frac{\pi}{\mathrm{2}}−{s}\right)= \\ $$$$=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({cos}\left({s}\right)\right){ds} \\ $$$$\Rightarrow\mathrm{2}{J}={J}+{J}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \left({ln}\left({sin}\left({x}\right)\right)+{ln}\left({cos}\left({x}\right)\right)\right){dx}= \\ $$$$=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({sin}\left({x}\right){cos}\left({x}\right)\right){dx}= \\ $$$$=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left(\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\right){dx}= \\ $$$$=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({sin}\left(\mathrm{2}{x}\right)\right){dx}−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\pi} {ln}\left({sin}\left({x}\right)\right){dx}−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\pi/\mathrm{2}} ^{\:\pi/\mathrm{2}} {ln}\left({cos}\left({x}\right)\right){dx}−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}}= \\ $$$$=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({cos}\left({x}\right)\right){dx}−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}}={J}−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\Rightarrow{J}=−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\Rightarrow{I}=−\mathrm{2}{J} \\ $$$$\Rightarrow\int_{\mathrm{2}} ^{\:\infty} \frac{\mathrm{8}{arcsec}\left({x}/\mathrm{2}\right)}{{x}^{\mathrm{3}} −\mathrm{4}{x}}{dx}=\pi{ln}\left(\mathrm{2}\right) \\ $$