Need help with a u-substitution problem

[Clandestiny]

Hey guys, so I'm trying to get a head start on some homework, and came across some problems I'm not sure how to solve.

here's an example:
integral of 3x(x^2 - 3)^7 dx.

I have u = x^2 - 3, with du = 2x.

My understanding is that du has to be in the integrand in some form. I can't make the 3 in the integrand work. One approach I've considered is dividing the 2x out, so it would be 1/2 du = x. Would it correct to have the integrand then look like this:

3/2 * u^7 * du ?

Which then would integrate to:

3/8(x^2-3)^8 + C


I'm not confident in my solution here, so any help would be greatly appreciated. Thanks guys!

 

[Deleted member 4993]

Hey guys, so I'm trying to get a head start on some homework, and came across some problems I'm not sure how to solve.

here's an example:
integral of 3x(x^2 - 3)^7 dx

I have u = x^2 - 3, with du = 2x.

My understanding is that du has to be in the integrand in some form. I can't make the 3 in the integrand work. One approach I've considered is dividing the 2x out, so it would be 1/2 du = x. Would it be correct to have the integrand then look like this:

3/2 * u^7 * du ?....... Correct

Which then would integrate to:

3/8(x^2-3)^8 + C

No... Almost....see below

I'm not confident in my solution here, so any help would be greatly appreciated. My area of confusion is in dealing with the term 3x. As 3x is a term, can the 3 be dealt with in the way I dealt with it? Thanks guys!

And apologies if this ended up getting double posted.

\(\displaystyle \displaystyle{\int \frac{3}{2}u^7 du }\)

\(\displaystyle \displaystyle{= \ \frac{3}{2}\frac{1}{8} u^8}\)

\(\displaystyle \displaystyle{= \ \frac{3}{16} u^8}\)

and continue...

There was no problem with the way dealt with '3' here....

 

[Clandestiny]

\(\displaystyle \displaystyle{\int \frac{3}{2}u^7 du }\)

\(\displaystyle \displaystyle{= \ \frac{3}{2}\frac{1}{8} u^8}\)

\(\displaystyle \displaystyle{= \ \frac{3}{16} u^8}\)

and continue...

There was no problem with the way dealt with '3' here....

Okay great, thank you so much! I see what I missed in the final integration, and that definitely cleared up my concerns with those pesky constants.

 

[lookagain]

I have u = x^2 - 3, with du = 2x.

Clandestiny,
don't forget the "dx" part.


du = (2x)dx

\(\displaystyle \displaystyle{\int \frac{3}{2}u^7 du }\)

\(\displaystyle \displaystyle{= \ \frac{3}{2}\frac{1}{8} u^8}\)

\(\displaystyle \displaystyle{= \ \frac{3}{16} u^8}\)

and continue...

\(\displaystyle \displaystyle{\int \frac{3}{2}u^7 du }\)

\(\displaystyle \displaystyle{= \ \bigg(\frac{3}{2}\bigg)\frac{1}{8} u^8} \ + \ C\)

\(\displaystyle \displaystyle{= \ \frac{3}{16} u^8} \ + \ C \)


And continue . . .

 

[Clandestiny]

Clandestiny,
don't forget the "dx" part.

du = (2x)dx

\(\displaystyle \displaystyle{\int \frac{3}{2}u^7 du }\)

\(\displaystyle \displaystyle{= \ \bigg(\frac{3}{2}\bigg)\frac{1}{8} u^8} \ + \ C\)

\(\displaystyle \displaystyle{= \ \frac{3}{16} u^8} \ + \ C \)

And continue . . .

Ha, thanks for the reminder! Now that we're getting into integration/substitution using trig identities (because calc 2 evidently moves crazy fast), I'm finally understanding the importance of that 'dx.' What I thought was a pesky formality turns out to be pretty necessary. Lesson learned; mathematic rules and procedures aren't arbitrary.