# 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:

$$\int x^9cos(x^5)$$

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

3. The Attempt at a Solution

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

Now I need to apply the substitution rule to find $$g(x)$$ by integrating $$cos(x^5)$$:
$$u=x^5$$
$$du=5x^4dx$$
Then maybe $$sin(u)du$$? 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

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tiny-tim
Homework Helper
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?

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
Homework Helper
Hint: $$x^4 \cos(x^5) = \frac{1}{5} \frac{\mathrm d}{\mathrm dx} \sin(x^5)$$.

Okaaay, here goes:

$$x^9sin(u)5x^4dx-\int sin(u)5x^4dx9x^8dx$$

No, you misunderstood the suggestions.

$$\int x^9\cos{x^5}dx=\int x^4x^5\cos{x^5}dx$$

$$u=x^5$$
$$du=5x^4dx$$

$$\frac 1 5\int u\cos udu$$

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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!

ahhha, ok. I'll be back!
Got it? You need to review if you're not getting it. This should be beyond you!!! First step is Calc 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.

$$\frac{1}{5}[u(sin(u))-(cos(u)+C)]$$ getting warmer?

Almost!!! Don't forget that ...

$$\int sin xdx=-\cos x+C$$

$$\frac 1 5 (x^5\sin x^5+\cos x^5)+C$$

Does it matter whether the C is inside or outside the brackets?

Gib Z
Homework Helper
No it doesn't. One fifth of an arbitrary real constant is still an arbitrary real constant isn't it.