Question

Integrals

Answer 1

\[\int\limits x^3\sqrt{9-x^2} dx\]

Answer 2

hint:
x^3=x^2*x
Use a substitution for the inside of that square root

Answer 3

i don't get it...

Answer 4

i would go with U sub

Answer 5

What I'm saying is u-sub the inside of that square root
and you need to notice you can also use that u-sub again for that x^2 part of x^3 thing

Answer 6

eh can we use trig here!?

Answer 7

trig isn't needed

Answer 8

Um I think i am supposed to use trig, if necessary

Answer 9

though i'm sure you could take that approach
the most simplest thing to do i think is to just algebraic sub (that you will use twice)

Answer 10

\[\int\limits_{}^{}x^2 \sqrt{9-x^2} x dx \]
If it helps write it like this first

Answer 11

looks to me trig is a good one here? why not?

Answer 12

make u=x^2 and use it, outside and inside of the squeare root by saying x^3 = x^2*x basically right?

Answer 13

let u=9-x^2
du=-2x dx

we said u=9-x^2
but that also means we can write x^2=9-u

Answer 14

replace the x^2's with (9-u)
replace x dx with -du/2

Answer 15

great approach @myininaya :)

Answer 16

but you can also trig it I'm sure

Answer 17

yeah think trig will complicate things. i just tried

Answer 18

long process...

Answer 19

yeah trig is a huge mess, i hate it

Answer 20

and thanks @kl0723
I'm just used to ones that sorta look like this one

Answer 21

what would be the answer so i can make sure i got this right

Answer 22

@willet I can walk you through both ways...

Have you tried doing it the first way I mentioned?

Answer 23

Yeah, if I can avoid the trig ill glady do it that way

Answer 24

so when you transformed the integral in terms of u
what did you get as your integral

Answer 25

\[-9/2 \int\limits -u \sqrt{u}\]

Answer 26

\[\int\limits_{}^{}x^3 \sqrt{9-x^2} dx =\int\limits_{}^{}x^2 \sqrt{9-x^2} x dx \]
you would let u=9-x^2 so x^2=9-u
so 2x dx=-du
so x dx=-du/2

So everywhere I see x^2 I will replace it with (9-u)
and wherever I see x dx i will replace it with -du/2
So we have this:
\[\int\limits_{}^{}(9-u)\sqrt{9-(9-u)} \frac{-1}{2} du \]

Answer 27

The inside of that square root simplifies to what you said which is u

Answer 28

\[\frac{-1}{2} \int\limits_{}^{}(9-u) \sqrt{u} du\]

Answer 29

Now write sqrt(u) as u^(1/2)
and distribute over the (9-u) part

Answer 30

\[\frac{-1}{2}\int\limits_{}^{}(9-u)u^\frac{1}{2}du \\ = \frac{-1}{2}\int\limits_{}^{}(9u^\frac{1}{2}-u^{1+\frac{1}{2}}) du \]
All I did was exactly what I told you to do
now the last thing couples you need to do is:
1) add 1 and 1/2
2) then integrate each term

Answer 31

you can integrate each term using the antiderivate (integration) power rule

Answer 32

ok i got \[6u ^{3/2}- \frac{ 2 }{ 5 }u^{5/2}\]

Answer 33

and that is without distributing that -1/2 multiple?

Answer 34

oops i forgot to do that, yes

Answer 35

Then also don't forget your plus C
and you need to finish up by replace u back in terms of x
using your substitution
u=9-x^2

Answer 36

alright i think i got it thanks

Answer 37

you can also integrate this by trig sub
it isn't too scary
did you least get the integral in terms of trig functions yet?

Answer 38

I think so, I got stuck with the trig

Answer 39

what did you the integral as in terms of trig functions?

Answer 40

ill send a picture of it, one sec. Im not sure if i did it right

Answer 41

sorry if you cant read my handwriting

Answer 42

I've been doing homework all day, so my mind is pretty much drained lol

Answer 43

6th line is where you went wrong
if x=3sin(theta)
then x^3=27sin^3(theta)

Answer 44

oh wow

Answer 45

integral should have been \[243 \int\limits_{}^{}\sin^3(\theta) \cos^2(\theta) d \theta \]

Answer 46

for this problem you would take the sin^3(theta) and write it as sin^2(theta) * sin(theta)
then rewrite sin^2(theta) as 1-cos^2(theta)
then distribute over the cos^2(theta) part
so you would \[243 \int\limits_{}^{} \sin^2(\theta) \cos^2(\theta) \sin(\theta) d \theta \\ = 243 \int\limits_{}^{}(1-\cos^2(\theta)) \cos^2(\theta) \sin(\theta) d \theta = \\ 243 \int\limits_{}^{}(\cos^2(\theta) -\cos^4(\theta)) \sin(\theta) d \theta \]
then you would make another sub
let v=cos(theta)

Answer 47

yeah you would do another Sub
involving trig
the end result will be the same though