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simplyfy
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how to get ln2 from that
@zepdrix
\[\int\limits_{\pi/6}^{\pi/2}\cot x dx\]
i know that cotx =ln sinx +c
The integral of cotx? Ok good c:
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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
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Oh you were doing something fancy, hmm
|dw:1361499527104:dw|
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