## ksaimouli Group Title simplyfy one year ago one year ago

1. ksaimouli

how to get ln2 from that

2. ksaimouli

@zepdrix

3. ksaimouli

$\int\limits_{\pi/6}^{\pi/2}\cot x dx$

4. ksaimouli

i know that cotx =ln sinx +c

5. zepdrix

The integral of cotx? Ok good c:

6. ksaimouli

which will givw me ln1-ln(1/2)

7. ksaimouli

0-ln(1)-ln(2) gives me -ln2 but the book gives ln 2

8. zepdrix

$\large \ln (a)-\ln(b)=\ln\left(\frac{a}{b}\right)$

9. zepdrix

where is the 0- coming from? :o

10. ksaimouli

ln(1)=0

11. zepdrix

Oh you were doing something fancy, hmm

12. ksaimouli

|dw:1361499527104:dw|

13. zepdrix

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?

14. ksaimouli

ohhhhhhh godddddddd

15. zepdrix

heh

16. ksaimouli

thx buddy

17. ksaimouli

|dw:1361500220284:dw|

18. ksaimouli

how to do this @zepdrix

19. zepdrix

Write it like this, you might be able to see your U substitution easier.$\large \int\limits (\ln x)^3 \left(\frac{1}{x}dx\right)$

20. zepdrix

Hmm no that wouldn't do much for us.

21. ksaimouli

lnx

22. zepdrix

Think about the derivative of natural log and also the derivative of 1/x. Maybe one of those pieces could be our u.

23. zepdrix

Does the derivative of either one show up in the integral?

24. ksaimouli

if i take derivative of lnx then i get 1/x hmmm

25. zepdrix

Hmmm interesting.

26. ksaimouli

no idea!

27. zepdrix

So if we let $$\large u=\ln x$$, taking the derivative of $$\large u$$ gives us, $$\large du=\dfrac{1}{x}dx$$ right?

28. ksaimouli

YES

29. zepdrix

Why is this in the biology section? lol

30. ksaimouli

oopsss i thought this is in math when i loged in i did not notice that

31. zepdrix

oh silly c:

32. zepdrix

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?

33. ksaimouli

ohk i got it