Question Number 201293 by sonukgindia last updated on 03/Dec/23
Answered by aleks041103 last updated on 03/Dec/23
![I=∫_2 ^( ∞) ((8arcsec(x/2)dx)/(x^3 −4x))= =∫_1 ^( ∞) ((8arcsec((2x)/2))/((2x)^3 −4(2x)))d(2x)=2∫_1 ^( ∞) ((arcsec(x))/(x(x^2 −1)))dx x=sec(t)⇒dx=sec(t)tan(t)dt ∫_0 ^( π/2) ((t sec(t) tan(t) dt)/(sec(t)(sec^2 (t)−1)))= =∫_0 ^( π/2) ((t cos(t))/(sin(t)))dt=∫_0 ^( π/2) td(ln(sin(t)))= =[tln(sin(t))]_0 ^(π/2) −∫_0 ^( π/2) ln(sin(t))dt =−∫_0 ^( π/2) ln(sin(t))dt J=∫_0 ^( π/2) ln(sin(t))dt=∫_(π/2) ^( 0) ln(sin((π/2)−s))d((π/2)−s)= =∫_0 ^( π/2) ln(cos(s))ds ⇒2J=J+J=∫_0 ^( π/2) (ln(sin(x))+ln(cos(x)))dx= =∫_0 ^( π/2) ln(sin(x)cos(x))dx= =∫_0 ^( π/2) ln(((sin(2x))/2))dx= =∫_0 ^( π/2) ln(sin(2x))dx−(π/2)ln(2)= =(1/2)∫_0 ^( π) ln(sin(x))dx−((πln(2))/2)= =(1/2)∫_(−π/2) ^( π/2) ln(cos(x))dx−((πln(2))/2)= =∫_0 ^( π/2) ln(cos(x))dx−((πln(2))/2)=J−((πln(2))/2) ⇒J=−((πln(2))/2) ⇒I=−2J ⇒∫_2 ^( ∞) ((8arcsec(x/2))/(x^3 −4x))dx=πln(2)](https://www.tinkutara.com/question/Q201306.png)
$${I}=\int_{\mathrm{2}} ^{\:\infty} \frac{\mathrm{8}{arcsec}\left({x}/\mathrm{2}\right){dx}}{{x}^{\mathrm{3}} −\mathrm{4}{x}}= \\ $$$$=\int_{\mathrm{1}} ^{\:\infty} \frac{\mathrm{8}{arcsec}\left(\left(\mathrm{2}{x}\right)/\mathrm{2}\right)}{\left(\mathrm{2}{x}\right)^{\mathrm{3}} −\mathrm{4}\left(\mathrm{2}{x}\right)}{d}\left(\mathrm{2}{x}\right)=\mathrm{2}\int_{\mathrm{1}} ^{\:\infty} \frac{{arcsec}\left({x}\right)}{{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{dx} \\ $$$${x}={sec}\left({t}\right)\Rightarrow{dx}={sec}\left({t}\right){tan}\left({t}\right){dt} \\ $$$$\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \frac{{t}\:{sec}\left({t}\right)\:{tan}\left({t}\right)\:{dt}}{{sec}\left({t}\right)\left({sec}^{\mathrm{2}} \left({t}\right)−\mathrm{1}\right)}= \\ $$$$=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \frac{{t}\:{cos}\left({t}\right)}{{sin}\left({t}\right)}{dt}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {td}\left({ln}\left({sin}\left({t}\right)\right)\right)= \\ $$$$=\left[{tln}\left({sin}\left({t}\right)\right)\right]_{\mathrm{0}} ^{\pi/\mathrm{2}} −\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({sin}\left({t}\right)\right){dt} \\ $$$$=−\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({sin}\left({t}\right)\right){dt} \\ $$$${J}=\int_{\mathrm{0}} ^{\:\:\pi/\mathrm{2}} {ln}\left({sin}\left({t}\right)\right){dt}=\int_{\pi/\mathrm{2}} ^{\:\:\mathrm{0}} {ln}\left({sin}\left(\frac{\pi}{\mathrm{2}}−{s}\right)\right){d}\left(\frac{\pi}{\mathrm{2}}−{s}\right)= \\ $$$$=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({cos}\left({s}\right)\right){ds} \\ $$$$\Rightarrow\mathrm{2}{J}={J}+{J}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \left({ln}\left({sin}\left({x}\right)\right)+{ln}\left({cos}\left({x}\right)\right)\right){dx}= \\ $$$$=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({sin}\left({x}\right){cos}\left({x}\right)\right){dx}= \\ $$$$=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left(\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\right){dx}= \\ $$$$=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({sin}\left(\mathrm{2}{x}\right)\right){dx}−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\pi} {ln}\left({sin}\left({x}\right)\right){dx}−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\pi/\mathrm{2}} ^{\:\pi/\mathrm{2}} {ln}\left({cos}\left({x}\right)\right){dx}−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}}= \\ $$$$=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({cos}\left({x}\right)\right){dx}−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}}={J}−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\Rightarrow{J}=−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\Rightarrow{I}=−\mathrm{2}{J} \\ $$$$\Rightarrow\int_{\mathrm{2}} ^{\:\infty} \frac{\mathrm{8}{arcsec}\left({x}/\mathrm{2}\right)}{{x}^{\mathrm{3}} −\mathrm{4}{x}}{dx}=\pi{ln}\left(\mathrm{2}\right) \\ $$