Menu Close

let-f-x-pi-6-pi-4-tant-2-x-cost-dt-with-x-real-1-determine-a-explicit-form-for-f-x-2-determine-also-g-x-pi-6-pi-4-tant-2-xcost-2-dx-3-find-the-value-of-pi-6-pi-4-

Tinku Tara 0 Comments
Pin




Question Number 72888 by mathmax by abdo last updated on 04/Nov/19
let f(x)=∫_(π/6) ^(π/4) ((tant)/(2+x cost))dt with x real 1)determine a explicit form for f(x) 2)determine also g(x)=∫_(π/6) ^(π/4) ((tant)/((2+xcost)^2 ))dx 3) find the value of ∫_(π/6) ^(π/4) ((tant)/((2+3cost)))dt and ∫_(π/6) ^(π/4) ((tant)/((2+3cost)^2 ))dt
$${let}\:{f}\left({x}\right)=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{tant}}{\mathrm{2}+{x}\:{cost}}{dt}\:\:{with}\:{x}\:{real} \\ $$$$\left.\mathrm{1}\right){determine}\:{a}\:{explicit}\:{form}\:{for}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right){determine}\:{also}\:{g}\left({x}\right)=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{tant}}{\left(\mathrm{2}+{xcost}\right)^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{tant}}{\left(\mathrm{2}+\mathrm{3}{cost}\right)}{dt}\:{and}\:\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{tant}}{\left(\mathrm{2}+\mathrm{3}{cost}\right)^{\mathrm{2}} }{dt} \\ $$
Commented by mathmax by abdo last updated on 05/Nov/19
we have f(x)=∫_(π/6) ^(π/4) ((sint)/(cost(2+xcost)))dt changent cost =u give f(x)=∫_((√3)/2) ^((√2)/2) ((−du)/(u(2+xu))) =−∫_((√3)/3) ^((√2)/2) (du/(u(xu +2))) let decompose F(u)=(1/(u(xu+2))) ⇒F(u)=(a/u) +(b/(xu +2)) a =(1/2) and b =lim_(u→−(2/x)) (xu+2)F(u)=(1/(−(2/x))) =−(x/2) ⇒ F(u)=(1/(2u))−(x/(2(xu+2))) ⇒f(x)=−(1/2)∫_((√3)/2) ^((√2)/2) ((1/u)−(x/(xu+2)))du =−(1/2)[ln∣(u/(xu+2))∣]_((√3)/2) ^((√2)/2) =−(1/2){ ln∣(((√2)/2)/(x((√2)/2)+2))∣−ln∣(((√3)/2)/(x((√3)/2)+2))∣} =−(1/2){ln∣((√2)/(4+x(√2)))∣−ln∣((√3)/(4+x(√3)))∣} =−(1/2){(1/2)ln(2)−ln∣4+x(√2)∣−(1/2)ln(3)+ln∣4+x(√3)∣} =−(1/4)(ln(2)−ln(3))+(1/2)ln∣((4+x(√2))/(4+x(√3)))∣ ⇒ f(x)=(1/4)ln((3/2))+(1/2)ln∣((4+x(√2))/(4+x(√3)))∣
$${we}\:{have}\:{f}\left({x}\right)=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{sint}}{{cost}\left(\mathrm{2}+{xcost}\right)}{dt}\:\:{changent}\:{cost}\:={u}\:{give} \\ $$$${f}\left({x}\right)=\int_{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} ^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} \:\:\:\frac{−{du}}{{u}\left(\mathrm{2}+{xu}\right)}\:=−\int_{\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}} ^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} \:\:\frac{{du}}{{u}\left({xu}\:+\mathrm{2}\right)}\:{let}\:{decompose} \\ $$$${F}\left({u}\right)=\frac{\mathrm{1}}{{u}\left({xu}+\mathrm{2}\right)}\:\Rightarrow{F}\left({u}\right)=\frac{{a}}{{u}}\:+\frac{{b}}{{xu}\:+\mathrm{2}} \\ $$$${a}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\:{and}\:{b}\:={lim}_{{u}\rightarrow−\frac{\mathrm{2}}{{x}}} \:\left({xu}+\mathrm{2}\right){F}\left({u}\right)=\frac{\mathrm{1}}{−\frac{\mathrm{2}}{{x}}}\:=−\frac{{x}}{\mathrm{2}}\:\Rightarrow \\ $$$${F}\left({u}\right)=\frac{\mathrm{1}}{\mathrm{2}{u}}−\frac{{x}}{\mathrm{2}\left({xu}+\mathrm{2}\right)}\:\Rightarrow{f}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} ^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} \left(\frac{\mathrm{1}}{{u}}−\frac{{x}}{{xu}+\mathrm{2}}\right){du} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\mid\frac{{u}}{{xu}+\mathrm{2}}\mid\right]_{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} ^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} =−\frac{\mathrm{1}}{\mathrm{2}}\left\{\:{ln}\mid\frac{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}{{x}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\mathrm{2}}\mid−{ln}\mid\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{{x}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\mathrm{2}}\mid\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\mid\frac{\sqrt{\mathrm{2}}}{\mathrm{4}+{x}\sqrt{\mathrm{2}}}\mid−{ln}\mid\frac{\sqrt{\mathrm{3}}}{\mathrm{4}+{x}\sqrt{\mathrm{3}}}\mid\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)−{ln}\mid\mathrm{4}+{x}\sqrt{\mathrm{2}}\mid−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{3}\right)+{ln}\mid\mathrm{4}+{x}\sqrt{\mathrm{3}}\mid\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\left({ln}\left(\mathrm{2}\right)−{ln}\left(\mathrm{3}\right)\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\mathrm{4}+{x}\sqrt{\mathrm{2}}}{\mathrm{4}+{x}\sqrt{\mathrm{3}}}\mid\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\mathrm{4}+{x}\sqrt{\mathrm{2}}}{\mathrm{4}+{x}\sqrt{\mathrm{3}}}\mid \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 05/Nov/19
2) we have f^′ (x) =−∫_(π/6) ^(π/4) ((cost×tant)/((2+xcost)^2 ))dt =−∫_(π/6) ^(π/4) ((sint)/((2+xcost)^2 ))dt =−g(x) ⇒g(x)=−f^′ (x) but f^′ (x) =(1/2)(((((4+x(√2))/(4+x(√3))))^′ )/((4+x(√2))/(4+x(√3)))) =((4+x(√3))/(2(4+x(√2))))×(((√2)(4+x(√3))−(√3)(4+x(√2)))/((4+x(√3))^2 )) =(1/(2(4+x(√2))(4+x(√3))))×4((√2)−(√3)) =((2(√2)−2(√3))/((4+x(√2))(4+x(√3)))) ⇒ g(x)=((2(√3)−2(√2))/((4+x(√2))(4+x(√3))))
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)\:=−\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{cost}×{tant}}{\left(\mathrm{2}+{xcost}\right)^{\mathrm{2}} }{dt}\:=−\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{sint}}{\left(\mathrm{2}+{xcost}\right)^{\mathrm{2}} }{dt} \\ $$$$=−{g}\left({x}\right)\:\Rightarrow{g}\left({x}\right)=−{f}^{'} \left({x}\right)\:\:{but}\: \\ $$$${f}^{'} \left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\frac{\left(\frac{\mathrm{4}+{x}\sqrt{\mathrm{2}}}{\mathrm{4}+{x}\sqrt{\mathrm{3}}}\right)^{'} }{\frac{\mathrm{4}+{x}\sqrt{\mathrm{2}}}{\mathrm{4}+{x}\sqrt{\mathrm{3}}}}\:=\frac{\mathrm{4}+{x}\sqrt{\mathrm{3}}}{\mathrm{2}\left(\mathrm{4}+{x}\sqrt{\mathrm{2}}\right)}×\frac{\sqrt{\mathrm{2}}\left(\mathrm{4}+{x}\sqrt{\mathrm{3}}\right)−\sqrt{\mathrm{3}}\left(\mathrm{4}+{x}\sqrt{\mathrm{2}}\right)}{\left(\mathrm{4}+{x}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{4}+{x}\sqrt{\mathrm{2}}\right)\left(\mathrm{4}+{x}\sqrt{\mathrm{3}}\right)}×\mathrm{4}\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right)\:=\frac{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{3}}}{\left(\mathrm{4}+{x}\sqrt{\mathrm{2}}\right)\left(\mathrm{4}+{x}\sqrt{\mathrm{3}}\right)}\:\Rightarrow \\ $$$${g}\left({x}\right)=\frac{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{2}\sqrt{\mathrm{2}}}{\left(\mathrm{4}+{x}\sqrt{\mathrm{2}}\right)\left(\mathrm{4}+{x}\sqrt{\mathrm{3}}\right)} \\ $$
Commented by mathmax by abdo last updated on 05/Nov/19
3)∫_(π/6) ^(π/4) ((tant)/((2+3cost)))dt =f(3)=(1/4)ln((3/2))+(1/2)ln∣((4+3(√2))/(4+3(√3)))∣
$$\left.\mathrm{3}\right)\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{tant}}{\left(\mathrm{2}+\mathrm{3}{cost}\right)}{dt}\:={f}\left(\mathrm{3}\right)=\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{4}+\mathrm{3}\sqrt{\mathrm{3}}}\mid \\ $$
Commented by mathmax by abdo last updated on 05/Nov/19
∫_(π/6) ^(π/4) ((sint)/((2+3cost)^2 ))dt =g(3)=((2(√3)−2(√2))/((4+3(√2))(4+3(√3))))
$$\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{sint}}{\left(\mathrm{2}+\mathrm{3}{cost}\right)^{\mathrm{2}} }{dt}\:={g}\left(\mathrm{3}\right)=\frac{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{2}\sqrt{\mathrm{2}}}{\left(\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}\right)\left(\mathrm{4}+\mathrm{3}\sqrt{\mathrm{3}}\right)} \\ $$
Commented by mathmax by abdo last updated on 05/Nov/19
sorry g(x)=∫_(π/6) ^(π/4) ((sint)/((2+xcost)^2 ))dt
$${sorry}\:{g}\left({x}\right)=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{sint}}{\left(\mathrm{2}+{xcost}\right)^{\mathrm{2}} }{dt} \\ $$
Answered by mind is power last updated on 04/Nov/19
f(x)=∫((sin(t)dt)/(cos(t)(2+xcos(t)) )) u=cos(t) f(x)=∫_((√3)/2) ^((√2)/2) ((−du)/(u(2+xu)))=∫_((√2)/2) ^((√3)/2) (du/(2u))−((xdu)/(2(2+xu)))=(1/2)ln(((√3)/( (√2))))−(1/2)ln(((2+((x(√3))/2))/(2+x((√2)/2)))) f′(x)=∫−((tan(t)cos(t))/((2+xcos(t))^2 ))dt =−∫((tan(t)(cos(t)+(2/x)−(2/x)))/((2+xcos(t))^2 ))=−(1/x)∫((tan(t))/((2+xcos(t))))+(2/x)∫((tan(t))/((2+xcos(t))^2 )) ⇒((xf′(x))/2)+((f(x))/2)=g(x) g(x)=(x/2){−((√3)/(4+x(√2)))+((√2)/(4+x(√2)))}+(1/2){ln(((√3)/( (√2))))−(1/2)ln(((4+x(√3))/(4+x(√2))))} ∫_(π/6) ^(π/4) ((tan(t))/((2+3cos(t))))dt=f(3)=(1/2)ln(((√3)/( (√2))))−(1/2)ln(((4+3(√3))/(4+3(√2)))) ∫_(π/6) ^(π/4) (((tan(t)))/((2+3cos(t))^2 ))dt=g(3)=(3/2)((((√2)−(√3))/(4+3(√2))))+((ln(((√3)/( (√2)))))/2)−(1/4)ln(((4+3(√3))/(4+3(√3))))
$$\mathrm{f}\left(\mathrm{x}\right)=\int\frac{\mathrm{sin}\left(\mathrm{t}\right)\mathrm{dt}}{\mathrm{cos}\left(\mathrm{t}\right)\left(\mathrm{2}+\mathrm{xcos}\left(\mathrm{t}\right)\right)\:} \\ $$$$\mathrm{u}=\mathrm{cos}\left(\mathrm{t}\right) \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\int_{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} ^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} \frac{−\mathrm{du}}{\mathrm{u}\left(\mathrm{2}+\mathrm{xu}\right)}=\int_{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \frac{\mathrm{du}}{\mathrm{2u}}−\frac{\mathrm{xdu}}{\mathrm{2}\left(\mathrm{2}+\mathrm{xu}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\frac{\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{2}}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{2}+\frac{\mathrm{x}\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{2}+\mathrm{x}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\right) \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\int−\frac{\mathrm{tan}\left(\mathrm{t}\right)\mathrm{cos}\left(\mathrm{t}\right)}{\left(\mathrm{2}+\mathrm{xcos}\left(\mathrm{t}\right)\right)^{\mathrm{2}} }\mathrm{dt} \\ $$$$=−\int\frac{\mathrm{tan}\left(\mathrm{t}\right)\left(\mathrm{cos}\left(\mathrm{t}\right)+\frac{\mathrm{2}}{\mathrm{x}}−\frac{\mathrm{2}}{\mathrm{x}}\right)}{\left(\mathrm{2}+\mathrm{xcos}\left(\mathrm{t}\right)\right)^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{x}}\int\frac{\mathrm{tan}\left(\mathrm{t}\right)}{\left(\mathrm{2}+\mathrm{xcos}\left(\mathrm{t}\right)\right)}+\frac{\mathrm{2}}{\mathrm{x}}\int\frac{\mathrm{tan}\left(\mathrm{t}\right)}{\left(\mathrm{2}+\mathrm{xcos}\left(\mathrm{t}\right)\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\mathrm{xf}'\left(\mathrm{x}\right)}{\mathrm{2}}+\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{2}}=\mathrm{g}\left(\mathrm{x}\right) \\ $$$$\mathrm{g}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\mathrm{2}}\left\{−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}+\mathrm{x}\sqrt{\mathrm{2}}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}+\mathrm{x}\sqrt{\mathrm{2}}}\right\}+\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{ln}\left(\frac{\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{2}}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{4}+\mathrm{x}\sqrt{\mathrm{3}}}{\mathrm{4}+\mathrm{x}\sqrt{\mathrm{2}}}\right)\right\} \\ $$$$\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{tan}\left(\mathrm{t}\right)}{\left(\mathrm{2}+\mathrm{3cos}\left(\mathrm{t}\right)\right)}\mathrm{dt}=\mathrm{f}\left(\mathrm{3}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\frac{\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{2}}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{4}+\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}}\right) \\ $$$$\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{4}}} \frac{\left(\mathrm{tan}\left(\mathrm{t}\right)\right)}{\left(\mathrm{2}+\mathrm{3cos}\left(\mathrm{t}\right)\right)^{\mathrm{2}} }\mathrm{dt}=\mathrm{g}\left(\mathrm{3}\right)=\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}}{\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}}\right)+\frac{\mathrm{ln}\left(\frac{\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{2}}}\right)}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\left(\frac{\mathrm{4}+\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}+\mathrm{3}\sqrt{\mathrm{3}}}\right) \\ $$$$ \\ $$
Commented by turbo msup by abdo last updated on 04/Nov/19
thsnk you sir.
$${thsnk}\:{you}\:{sir}. \\ $$
Commented by mind is power last updated on 04/Nov/19
y,re welcom
$$\mathrm{y},\mathrm{re}\:\mathrm{welcom} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *