Question

Solving indefinite integrals with substitution: What do I do when there is a term outside of my u group that is a higher power than my du. Can I square or cube my du? And if so what happens after that?

Answer 1

Question is \[\int\limits 18x^3(3x^2+1)^{1/2} dx\]

Answer 2

Yes, du doesn't really work like that, you have some misconception that I'll try to help you figure out.

Answer 3

Ahhh I see what's going on. You can't do this substitution, you need to find another way.

Answer 4

Oh really? Greeeeaat. But thank you for the help!

Answer 5

You should try using 3x^2+1=u. I think that'll work.

Answer 6

It does. =D

Answer 7

You have to be clever though, so just think about it for a second and hopefully you notice something to make this substitution work for you. ;D

Answer 8

What do I need to do to eliminate the 18x^3?

Answer 9

Alright so u=3x^2-1 right? What is dx? I'll help walk you through it.

Answer 10

u = 3x^2 + 1
du/dx u = 3x^2 + 1
du/dx = 6x
du = 6xdx
du/6x = dx

do the substitution, and you will see

Answer 11

but kainul can explain it to you I guess

Answer 12

Yeah, it's whatever lol, it's a pretty fun integral honestly.

Answer 13

We have differing ideas of fun. My prof just taught us substitution in 20 mins today and our final is next week heavily based on it.

Answer 14

Well if you take cal 2, u sub will be one of your best friends/most used tool in math toolkit next to trig sub and integration by parts. =D

Answer 15

I could totally just see myself solving this with a substitution then just doing by parts twice

Answer 16

but I'm rusty

Answer 17

By parts? I don't think you need more than u-substitution here.

Answer 18

yeah probably not

Answer 19

Ifstretc have you figured it out yet or still working on it and need help?

Answer 20

So I may have done it wrong, but putting dx=du/6x into the equation (as well as the u, thats easy) i end up with 3x^2(u)^(1/2)du

Answer 21

Exactly! Now look back at what you chose for your u. Notice how it has x^2 in there?

Answer 22

oooooooh

Answer 23

Yeah, see what I mean by fun?

Answer 24

But +1?

Answer 25

Subtract 1 from each side.

Answer 26

Don't forget you still have the power of algebra by your side in the war against calculus.

Answer 27

Wow I need sleep I cant believe I didnt see that

Answer 28

\[\int\limits_{}^{}(u-1)u^{1/2}du\]

Answer 29

(u - 1) = 3x^2

Answer 30

Okay so now i have that. so take the integral of that...

Answer 31

now to solve just multiply u^(1/2) into (u-1)

Answer 32

Distribute out the u^(1/2) and you basically have a regular old polynomial to integrate. No problems?

Answer 33

and split the integral

Answer 34

so\[u ^{3/2}-u ^{1/2}du\]

Answer 35

yeah looks good

Answer 36

now split it and take the antiderivative of each

Answer 37

and dont forget + c

Answer 38

And plug all your stuff back in terms of x, that's also important =P

Answer 39

Okay perfect. You two both deserve a huge thank you for walking my calc challenged self through that!

Answer 40

Haha no problem, I think it was fun for both of us. Calculus gets trickier and more clever the further on you go. I get excited sometimes when I see hard problems now because they're so satisfying to figure out.

Answer 41

Fortunately for myself this will be my only calculus class in uni. But it has been challenging and at times rewarding :)

Answer 42

this problem wasn't that hard kainui, it can be solved without really thinking using a method I stated above but it would have been long and tedious

Answer 43

This one is so good at misleading you straight to the 3x^2 from x^3, they're kind of jerks for that lol.

Answer 44

but yeah this was a tricky problem :)

Answer 45

wait no I'm wrong lol

Answer 46

I suppose, but I don't think integration by parts was taught to me when I was in cal 1, sounded like he just learned u-substitution. I can imagine the u's and dv's and stuff

Answer 47

I was taught all major integration techniques in calc 1 :\ like 50% of my exam was integration

Answer 48

We've only just skimmed it in our last week of calc. There was a lot of focus on curve sketching in my course

Answer 49

Yeah I dont think my method would work actually so disregard that

Answer 50

Damn, that sucks. I think the point where I decided I liked calculus a lot was when we started revolving functions around axes and finding the volume. It's a lot of dividing by 0 and multiplying by infinitely small things and adding infinite numbers together that gets me into it honestly lol.

Answer 51

Sounds interesting, but not for me. :)

Answer 52

I really have to wonder about how crazily smart the people who invented all this stuff were. It's kind of like getting a glimpse of what it's like to be a genius because you get to use their mental problem solving models. It's kind of weird and fascinating. Anyways I can talk about math all night long... So good luck, I'm going to bed.

Answer 53

Thank you again!

Answer 54

you can write \[\large \int_{}{}18x^3(3x^2+1)^{1/2}dx=\int_{}{}18x^2(3x^2+1)^{1/2}(xdx)\]