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how to get ln2 from that
\[\int\limits_{\pi/6}^{\pi/2}\cot x dx\]

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Other answers:

i know that cotx =ln sinx +c
The integral of cotx? Ok good c:
which will givw me ln1-ln(1/2)
0-ln(1)-ln(2) gives me -ln2 but the book gives ln 2
\[\large \ln (a)-\ln(b)=\ln\left(\frac{a}{b}\right)\]
where is the 0- coming from? :o
ln(1)=0
Oh you were doing something fancy, hmm
|dw:1361499527104:dw|
Ok i see what you're doing. You just missed a negative in the middle.\[\large 0-\ln(1/2) \qquad = \qquad 0-\left(\ln1-\ln2\right)\]See where you missed it?
ohhhhhhh godddddddd
heh
thx buddy
|dw:1361500220284:dw|
how to do this @zepdrix
Write it like this, you might be able to see your `U substitution` easier.\[\large \int\limits (\ln x)^3 \left(\frac{1}{x}dx\right)\]
Hmm no that wouldn't do much for us.
lnx
Think about the derivative of natural log and also the derivative of 1/x. Maybe one of those pieces could be our u.
Does the derivative of either one show up in the integral?
if i take derivative of lnx then i get 1/x hmmm
Hmmm interesting.
no idea!
So if we let \(\large u=\ln x\), taking the derivative of \(\large u\) gives us, \(\large du=\dfrac{1}{x}dx\) right?
YES
Why is this in the biology section? lol
oopsss i thought this is in math when i loged in i did not notice that
oh silly c:
So that would be the correctly substitution, because we have a suitable \(u\) and \(du\) that we can plug in. Understand how to plug them in?
ohk i got it

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