anonymous 5 years ago calculus question is there a typo here,

1. anonymous

2. anonymous

number 2, where it says then dz = f(z) du, shouldn't it say, then du = f(z) dz

3. anonymous

No, it's not a typo.

4. anonymous

that doesnt make sense

5. anonymous

if u = sin z , right? then du = cos z dz ,

6. anonymous

Right. And dz =?

7. anonymous

dz = 1/ cos z du ?

8. anonymous

yes

9. anonymous

why would they ask that , that seems strange

10. anonymous

Not really, that's how I solve all my u-subs.

11. anonymous

hmmm

12. anonymous

thats redundant

13. anonymous

Suit yourself. ;p

14. anonymous

one sec, let me check

15. anonymous

where exactly are you doing u substitution then

16. anonymous

u = sin z , then du = cos z dz , so if dz = du/ cos z we have integral du/ u^4, oh the cos z cancels/

17. anonymous

you might as well just substitute it directly

18. anonymous

so you get integral du / u^4

19. anonymous

this saves the step of having to plug that dz back in and then cancel, the whole point of u substitution i thought was to change the variables

20. anonymous

$u = sin\ z \implies du = (cos\ z)dz \implies dz = \frac{1}{cos\ z} du$ $\implies \int \frac{cos\ z}{sin^4z}dz = \int \frac{cos\ z}{u^4}(\frac{1}{cos\ z})du$ $= \int \frac{1}{u^4} du$

21. anonymous

Sometimes it works nicely, othertimes you have to do a bit of finagling.

22. anonymous

I'm not sure what you mean by redundancy.

23. anonymous

its redundant, you can skip that step , ok because du = cos z dz, and you already have cos z dz in the numerator

24. anonymous

Sure, but sometimes it's not that obvious.

25. anonymous

what do you mean

26. anonymous

i NEVER use this approach

27. anonymous

I always do.

28. anonymous

give me problem and i will show yuo my approach

29. anonymous

I understand the other way, but too often it's just a matter of doing something in your head and I prefer to have it spelled out explicitly.

30. anonymous

hmmm, it seems odd , i guess i learned it a different way

31. anonymous

everyone's brains work a little differently =)

32. anonymous

well let me show you what i do, and you compare it with what you do

33. anonymous

give me a problem, lets see

34. anonymous

integral x e^(x^2) dx

35. anonymous

u = x^2 , du = 2x dx , du/2 = x dx (its ok to divide out constant) so integral e^u du / 2

36. anonymous

sure, fine.

37. anonymous

ok :)

38. anonymous