Question Number 110875 by bemath last updated on 31/Aug/20
$$\left(\mathrm{1}\right)\underset{\mathrm{e}} {\overset{\mathrm{e}^{\mathrm{e}} } {\int}}\:\frac{\mathrm{ln}\:\left(\mathrm{x}\right).\mathrm{ln}\:\left(\mathrm{ln}\:\left(\mathrm{x}\right)\right)}{\mathrm{x}}\:\mathrm{dx}\:? \\ $$$$\left(\mathrm{2}\right)\underset{{x}\rightarrow\pi/\mathrm{4}} {\mathrm{lim}}\:\frac{\mathrm{cosec}\:^{\mathrm{2}} \mathrm{x}−\mathrm{2}}{\mathrm{cot}\:\mathrm{x}−\mathrm{1}} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Given}\:\begin{cases}{\mathrm{xy}=\frac{\mathrm{16y}−\mathrm{9x}}{\mathrm{45}}}\\{\frac{\mathrm{4}}{\:\sqrt{\mathrm{x}}}−\frac{\mathrm{3}}{\:\sqrt{\mathrm{y}}}\:=\:\mathrm{5}}\end{cases} \\ $$$$\Rightarrow\mathrm{find}\:\mathrm{9}\sqrt{\mathrm{xy}} \\ $$
Answered by Dwaipayan Shikari last updated on 31/Aug/20
![∫_e ^e^e ((log(x)log(log(x)))/x)dx (logx=t,(1/x)=(dt/dx)) ∫_1 ^e tlog(t)dt [logt.(t^2 /2)]_1 ^e −∫_1 ^e (t/2)=(e^2 /2)−(e^2 /4)+(1/4)=(1/4)(e^2 +1)](https://www.tinkutara.com/question/Q110882.png)
$$\int_{{e}} ^{{e}^{{e}} } \frac{{log}\left({x}\right){log}\left({log}\left({x}\right)\right)}{{x}}{dx}\:\:\:\left({logx}={t},\frac{\mathrm{1}}{{x}}=\frac{{dt}}{{dx}}\right) \\ $$$$\int_{\mathrm{1}} ^{{e}} {tlog}\left({t}\right){dt} \\ $$$$\left[{logt}.\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{1}} ^{{e}} −\int_{\mathrm{1}} ^{{e}} \frac{{t}}{\mathrm{2}}=\frac{{e}^{\mathrm{2}} }{\mathrm{2}}−\frac{{e}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}}\left({e}^{\mathrm{2}} +\mathrm{1}\right) \\ $$
Commented by bemath last updated on 31/Aug/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}.\: \\ $$
Commented by bemath last updated on 31/Aug/20
Answered by Dwaipayan Shikari last updated on 31/Aug/20
$$\left.\mathrm{2}\right)\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} {x}}{{sin}^{\mathrm{2}} {x}\left({cosx}−{sinx}\right)}.{sinx}=\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\frac{{cos}^{\mathrm{2}} {x}−{sin}^{\mathrm{2}} {x}}{{sinx}\left({cosx}−{sinx}\right)}=\sqrt{\mathrm{2}}.\left({sinx}+{cosx}\right) \\ $$$$=\mathrm{2} \\ $$
Commented by bemath last updated on 31/Aug/20
Answered by bobhans last updated on 31/Aug/20
$$\left(\mathrm{3}\right)\:\begin{cases}{\mathrm{45}=\frac{\mathrm{16y}−\mathrm{9x}}{\mathrm{xy}}=\frac{\mathrm{16}}{\mathrm{x}}−\frac{\mathrm{9}}{\mathrm{9}}=\left(\frac{\mathrm{4}}{\:\sqrt{\mathrm{x}}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{y}}}\right)^{\mathrm{2}} }\\{\frac{\mathrm{4}}{\:\sqrt{\mathrm{x}}}−\frac{\mathrm{3}}{\:\sqrt{\mathrm{y}}}\:=\:\mathrm{5}}\end{cases} \\ $$$$\begin{cases}{\mathrm{45}=\left(\frac{\mathrm{4}}{\:\sqrt{\mathrm{x}}}−\frac{\mathrm{3}}{\:\sqrt{\mathrm{y}}}\right)\left(\frac{\mathrm{4}}{\:\sqrt{\mathrm{x}}}+\frac{\mathrm{3}}{\:\sqrt{\mathrm{y}}}\right)…\left(\mathrm{i}\right)}\\{\mathrm{5}=\frac{\mathrm{4}}{\:\sqrt{\mathrm{x}}}−\frac{\mathrm{3}}{\:\sqrt{\mathrm{y}}}…\left(\mathrm{ii}\right)}\end{cases} \\ $$$$\Leftrightarrow\:\mathrm{45}\:=\:\mathrm{5}\left(\frac{\mathrm{4}}{\:\sqrt{\mathrm{x}}}+\frac{\mathrm{3}}{\:\sqrt{\mathrm{y}}}\right)\:;\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{x}}}+\frac{\mathrm{3}}{\:\sqrt{\mathrm{y}}}\:=\:\mathrm{9}…\left(\mathrm{iii}\right) \\ $$$$\left(\mathrm{i}\right)+\left(\mathrm{iii}\right)\Rightarrow\:\frac{\mathrm{8}}{\:\sqrt{\mathrm{x}}}\:=\:\mathrm{14}\Rightarrow\sqrt{\mathrm{x}}\:=\:\frac{\mathrm{4}}{\mathrm{7}}\:\mathrm{and}\:\frac{\mathrm{3}}{\:\sqrt{\mathrm{y}}}=\mathrm{9}−\mathrm{7} \\ $$$$\sqrt{\mathrm{y}}\:=\:\frac{\mathrm{3}}{\mathrm{2}}.\:\mathrm{therefore}\:\mathrm{9}\sqrt{\mathrm{xy}}\:=\:\mathrm{9}×\frac{\mathrm{4}}{\mathrm{7}}×\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$=\:\frac{\mathrm{54}}{\mathrm{7}} \\ $$$$ \\ $$
Commented by bemath last updated on 31/Aug/20
$$\mathrm{thank}\:\mathrm{you} \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 31/Aug/20
![I =∫_e ^e^e ((ln(x)ln(lnx))/x) dx changement lnx =t give I =∫_1 ^e ((t ln(t))/e^t ) e^t dt =∫_1 ^e tln(t)dt =[(t^2 /2)ln(t)]_1 ^e −∫_1 ^e (t^2 /2)(dt/t) =(e^2 /2) −(1/2)∫_1 ^e t dt =(e^2 /2)−(1/2)[(t^2 /2)]_1 ^e =(e^2 /2)−(1/4){e^2 −1} =(e^2 /4) +(1/4) ⇒ I =((1+e^2 )/4)](https://www.tinkutara.com/question/Q110938.png)
$$\mathrm{I}\:=\int_{\mathrm{e}} ^{\mathrm{e}^{\mathrm{e}} } \:\frac{\mathrm{ln}\left(\mathrm{x}\right)\mathrm{ln}\left(\mathrm{lnx}\right)}{\mathrm{x}}\:\mathrm{dx}\:\:\mathrm{changement}\:\mathrm{lnx}\:=\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{I}\:=\int_{\mathrm{1}} ^{\mathrm{e}} \:\frac{\mathrm{t}\:\mathrm{ln}\left(\mathrm{t}\right)}{\mathrm{e}^{\mathrm{t}} }\:\mathrm{e}^{\mathrm{t}} \:\mathrm{dt}\:=\int_{\mathrm{1}} ^{\mathrm{e}} \:\mathrm{tln}\left(\mathrm{t}\right)\mathrm{dt}\:=\left[\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}\mathrm{ln}\left(\mathrm{t}\right)\right]_{\mathrm{1}} ^{\mathrm{e}} −\int_{\mathrm{1}} ^{\mathrm{e}} \frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}\frac{\mathrm{dt}}{\mathrm{t}} \\ $$$$=\frac{\mathrm{e}^{\mathrm{2}} }{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{e}} \:\mathrm{t}\:\mathrm{dt}\:=\frac{\mathrm{e}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{1}} ^{\mathrm{e}} \:=\frac{\mathrm{e}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\left\{\mathrm{e}^{\mathrm{2}} −\mathrm{1}\right\}\:=\frac{\mathrm{e}^{\mathrm{2}} }{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow \\ $$$$\mathrm{I}\:=\frac{\mathrm{1}+\mathrm{e}^{\mathrm{2}} }{\mathrm{4}} \\ $$