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Lammy Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{1}^{e}1/(x \sqrt{1+(lnx)^2}) dx\]
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
\[\int\limits_{1}^{e}\frac{dx}{x \sqrt{1+(\ln(x))^2}}\] \[\text{ Let } a=\ln(x) => da=\frac{dx}{x}\] if x=e then a=ln(e)=1 if x=1 then a=ln(1)=0 So we have \[\int\limits_{0}^{1}\frac{da}{\sqrt{1+a^2}} \]
 2 years ago

shaan_iitk Group TitleBest ResponseYou've already chosen the best response.0
ln x= t .. then this becomes dt/(sqrt(1+t^2)) .. now put t = tany and you get secydy = ln(secy + tany)
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
\[\text{ Let } a=\tan(\theta) => da=\sec^2(\theta) d \theta\] if a=1 then theta=arctan(1)=pi/4 if a=0 then theta=arctan(0)=0 \[\int\limits_{0}^{\frac{\pi}{4}}\frac{\sec^2(\theta) d \theta}{\sqrt{1+\tan^2(\theta)}}\]
 2 years ago

Lammy Group TitleBest ResponseYou've already chosen the best response.0
yea thats as far as i got and lost
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
\[\int\limits_{0}^{\frac{\pi}{4}}\sec(\theta) d \theta \text{ (note :} 1+\tan^2(\theta) =\sec^2(\theta) \text{ )}\]
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
\[=\ln\sec(\theta)+\tan(\theta)_0^\frac{\pi}{4}=\ln\sec(\frac{\pi}{4})+\tan(\frac{\pi}{4})\ln\sec(0)+\tan(0)\]
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
\[=\ln(\frac{1}{\frac{\sqrt{2}}{2}}+\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}})\ln(\frac{1}{1}+\frac{0}{1})\]
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
\[=\ln(\frac{2}{\sqrt{2}}+1)\ln(1)=\ln(\frac{2 \sqrt{2}}{2}+1)0=\ln(\sqrt{2}+1)\]
 2 years ago

Lammy Group TitleBest ResponseYou've already chosen the best response.0
for the absolute values, how do you get rid of it?
 2 years ago

Lammy Group TitleBest ResponseYou've already chosen the best response.0
thanks for your help! very clear instruction. i always liked your help =)
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
the number inside was positive you don't need them anymore
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
\[\ln1=\ln(1)\] You don't need the absolute value since 1=1
 2 years ago
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