## AmTran_Bus one year ago Can someone help me with this integration by parts?

1. AmTran_Bus

|dw:1433873040803:dw|

2. AmTran_Bus

|dw:1433873069974:dw|

3. AmTran_Bus

That is supposed to say du= 2r dr

4. anonymous

you forgot the sqrt around the "4+r^2" actually i dont think you need int by parts here .... just the substitution --> u = 4+r^2 du = 2r dr

5. AmTran_Bus

Hum. This problem is from the integration by parts chapter in the Stewart book :(

6. AmTran_Bus

But ok. Lets do it your way! Can you guide me through it?

7. anonymous

sure ... double check you copied the question correctly though

8. AmTran_Bus

Ok. I did.

9. anonymous

after substitution you get $\frac{(u-4) r}{\sqrt{u}} * \frac{du}{2r}$

10. anonymous

the "r" cancels leaving $\frac{1}{2} \int\limits \frac{u-4}{\sqrt{u}} du$

11. AmTran_Bus

Ok, can you explain the after substitution step? Is u=4+r^2?

12. anonymous

yes so since u = 4 +r^2 , if you solve for r^2 you get r^2 = u-4 --> r^3 = r^2 *r = (u-4)r

13. AmTran_Bus

Got ya. Thanks.

14. AmTran_Bus

Don't we have to solve for out limits?

15. AmTran_Bus

*our

16. anonymous

yes but first you must find integral .... then plug in limits at the end

17. AmTran_Bus

18. anonymous

can you finish the integral? $\frac{u-4}{\sqrt{u}} = u^{1/2} -4u^{-1/2}$

19. phi

FYI if you want to use integration by parts, notice that to integrate $(4+r^2)^{-\frac{1}{2} }$ Taking that as a hint, write the problem as $\int \frac{r^3}{\sqrt{4+r^2}} dr = \int r^2 (4+r^2)\ r\ dr$ and let $u= r^2 \\ dv = (4+r^2)\ r\ dr$

20. AmTran_Bus

Ok. So the (4+r^2)^(-1/2) was basically put on top now, right?

21. phi

yes, only because I don't like square roots.

22. phi

to integrate $(4+r^2)^{-\frac{1}{2} }$ we need 2 r dr

23. AmTran_Bus

So we havae|dw:1433877516461:dw|

24. AmTran_Bus

Im just trying to get to what you did to be sure I get it.

25. phi

so split the r^3 into r^2 and r (and put the r next to the dr)

26. AmTran_Bus

Ok. r^2 *r

27. phi

yes. so it should look like this $\int \frac{r^3}{\sqrt{4+r^2}} dr = \int r^2 (4+r^2)^{-\frac{1}{2}}\ r\ dr$

28. AmTran_Bus

Very nice. So now I can just do the regular U and dv stuff.

29. phi

yes, u= r^2 dv = what is left over

30. AmTran_Bus

Thank you so very much. I still have just a small issue with how you got|dw:1433877820455:dw|

31. AmTran_Bus

Sorry for that mess. And for asking you so much.

32. AmTran_Bus

I get how you brought the sqrt up top.

33. AmTran_Bus

|dw:1433877903541:dw|

34. phi

you don't have to change the square root to an exponent, but I find it easier to think about|dw:1433877934816:dw|

35. phi

$\frac{1}{x}= x^{-1}$ $\sqrt{x} = x^\frac{1}{2}$ $\frac{1}{\sqrt{x}}=\frac{1}{x^\frac{1}{2}}=x^{-\frac{1}{2}}$

36. AmTran_Bus

Thanks so much phi. I am really glad you showed me how to do it this way.

37. phi

You still have to do the integral

38. AmTran_Bus

I know, from all the U= and DV= stuff. But I'm pretty good with that.