Question

Symbolically evaluate the given integrals. All work must be shown and proper notation must be used.

(x+1)^2*(cos(x^3+3x^2+3x))dx

Evaluated from 1 to 4

Answer 1

\[1\div( \sqrt{x}(\sqrt{x+1)}^{2}\]

Answer 2

for real?

Answer 3

is it possible that you can use a u - sub? lets check

Answer 4

Let \(x^3+3x^2 + 3x = u \)
\(du = 3x^2+6x+3=3(x^2+2x+1 \, dx)\)

Answer 5

\((x+1)^2=x^2+2x+1\) and
\[\frac{d}{dx}[x^3+3x^2+3x]=3x^2+6x+3\] how about that !!

Answer 6

oh what @kirbykirby said

Answer 7

i ended up finding it to be (sin(x^3)+3x^2+3x))/3

but i found before that the integral becomes (1/3)*\[(1/3)*\int\limits_{\cos(u)du}\]

Answer 8

@satellite73 @kirbykirby and honestly idk where the 1/3 came from..

Answer 9

thats what im trying to figure out where it came from

Answer 10

You should have a 1/3 factor because when you got: \( du =3(x^2+2x+1 )dx\) This implies:
\[\frac{1}{3}du=x^2+2x+1 \, dx \], so in your substitution for (x+1)^2, you see there is a 1/3 factor. However I am not sure why you have 1/3 in front of the integral, and also a divided by 3 within the integrand.

Answer 11

Ah ok that makes more sense but why would I need to factor? I just divided by three combining the (1/3) with the answer to make it simpler.

Answer 12

@kirbykirby

Answer 13

do you mean why I factored out 3 from \(3x^2+6x+3 \,dx\) to get \(3(x^2+2x+1)\, dx\) ?

Answer 14

yes @kirbykirby

Answer 15

well just so it would look more obvious for your substitution, since there was \((x+1)^2 = x^2+2x+1\) in your integral.

Answer 16

wait so your du would always end up being somehow what was actually missing or not used in the unevaluated integral? like say u didnt use (x+1)^2 and so somehow du is just supposed to be what was not used(x+1)^2? or does it just end up being like that?

Answer 17

just end up coincidentally representing (x+1)^2?

Answer 18

I'm not 100% sure I fully understand the question, but I'll do my best to answer.
Your integral had the factor \((x+1)^2\) in front, and there was \(x^3+3x^2+3x\) inside the cosine. So u-substitution tries to do a substitution in your integral where you can find one part of your integral that is the derivative of your substitution \(u\). It's important that all of the integral gets "used up" so that all the variables go from \(x\) to \(u\), with the exception of coefficients. I don't know if that sort of answers the question.

In fact, it is not a coincidence that they made the integral look that way so that you could use u-substitution. It was designed that way with the factor \((x+1)^2\) so that you can use that technique. Obviously, not all integrals will be this convenient! But that's when you will discover more integration techniques for different/harder integrals!

Answer 19

that makes perfect sense and clears it up that would greatly explain why I would need to factor and really helps me for future integration thanks @kirbykirby !

Answer 20

your welcome :)