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Substitution for indef. integral

1. Homework Statement
Hey folks, I think I know how to solve this by parts but I need a substitution to get there. I've been staring at examples for a while but I still don't understand how to apply the substitution rule. Anyway, here's the integral:

[tex]\int x^9cos(x^5)[/tex]

2. Homework Equations
integration by parts: [tex]\int f(x)g\prime (x)dx=f(x)g(x)-\int g(x)f\prime (x)dx[/tex]
substitution rule: [tex]\int f(g(x))g\prime (x)dx=\int f(u)du[/tex]

3. The Attempt at a Solution

By applying integration by parts:
[tex]f(x)=x^9[/tex]
[tex]f\prime (x)=9x^{8}[/tex]
[tex]g\prime (x)=cos(x^5)[/tex]

Now I need to apply the substitution rule to find [tex]g(x)[/tex] by integrating [tex]cos(x^5)[/tex]:
[tex]u=x^5[/tex]
[tex]du=5x^4dx[/tex]
Then maybe [tex]sin(u)du[/tex]? I'm not sure how to proceed. Thanks!
I've got another post that's still unanswered just in case you've got some more time to kill :wink:
 
Last edited:

Answers and Replies

tiny-tim
Science Advisor
Homework Helper
25,789
249
Nooo … if you're going to use substitution, do the substitution first!

(If you do the integrating by parts first, then split it differently - the object is to make it easier!)

As you said, u=x^5 is a good idea … but it's du=5x^4dx.

Then integrate by parts!

What do you get? :smile:
 
222
0
It;s simple , just break x^9 into x^5*x^4 , then take X^5 = t 5x^4dx=dt , therefore integral becomes 5 t cos(t) dt , solve using by parts
 
CompuChip
Science Advisor
Homework Helper
4,284
47
Hint: [tex]x^4 \cos(x^5) = \frac{1}{5} \frac{\mathrm d}{\mathrm dx} \sin(x^5)[/tex].
 
Okaaay, here goes:

[tex]x^9sin(u)5x^4dx-\int sin(u)5x^4dx9x^8dx[/tex]
 
1,750
1
No, you misunderstood the suggestions.

[tex]\int x^9\cos{x^5}dx=\int x^4x^5\cos{x^5}dx[/tex]

[tex]u=x^5[/tex]
[tex]du=5x^4dx[/tex]

[tex]\frac 1 5\int u\cos udu[/tex]
 
Last edited:
1,750
1
You're doing a substitution first!!! Then you will apply Parts.

u=x^5, so replace all x^5 with u
du=5x^4dx, so x^4dx appears in your original Integral, so that goes away.
 
ahhha, ok. I'll be back!
 
1,750
1
Yeah, I did cal 1 years ago so I'm sort of reviewing points from it as I slam up against them. I have been looking at substitution examples for a while but I need to grind through a few problems to shake the rust off. I think this one is coming along now though. I'll be back with an answer.
 
[tex]\frac{1}{5}[u(sin(u))-(cos(u)+C)][/tex] getting warmer?
 
1,750
1
Almost!!! Don't forget that ...

[tex]\int sin xdx=-\cos x+C[/tex]

[tex]\frac 1 5 (x^5\sin x^5+\cos x^5)+C[/tex]
 
Does it matter whether the C is inside or outside the brackets?
 
Gib Z
Homework Helper
3,344
4
No it doesn't. One fifth of an arbitrary real constant is still an arbitrary real constant isn't it.
 

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