Substitution for indef. integral

[silicon_hobo]

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:

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

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: $$\int f(g(x))g\prime (x)dx=\int f(u)du$$

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

 

[tiny-tim]

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?

 

[mkbh_10]

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]

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

 

[silicon_hobo]

Okaaay, here goes:

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

 

[rocomath]

No, you misunderstood the suggestions.

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

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

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

 

[rocomath]

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.

 

[silicon_hobo]

ahhha, ok. I'll be back!

 

[rocomath]

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!

 

[silicon_hobo]

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.

 

[silicon_hobo]

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

 

[rocomath]

Almost! Don't forget that ...

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

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

 

[silicon_hobo]

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

 

[Gib Z]

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