1. ## Integration by Parts

Hello,

I'm having trouble on a fairly basic integration by parts question:

The Integral of: 27x^2/cos(3x)

I remember going over something about using U substitution on cos(3x) but I can't remember how to do it.

Any help would be appreciated.

2. Originally Posted by Smilie
Hello,

I'm having trouble on a fairly basic integration by parts question:

The Integral of: 27x^2/cos(3x)

I remember going over something about using U substitution on cos(3x) but I can't remember how to do it.

Any help would be appreciated.
I'd start with the obvious: Let $u = 3x$. Then do your integration by parts.

-Dan

3. Originally Posted by Smilie
Hello,

The Integral of: 27x^2/cos(3x)
Maybe this is a "wrong" integral?

4. Originally Posted by curvature
Maybe this is a "wrong" integral?
Certainly looks that way:

5. Maybe the integrand is this: $\frac{27x^2}{cos(x^3)}$. That makes the substitution much easier.

6. ## My mistake

Ah, sorry I typed the wrong integral X_X. Anyways I was able to get the answer and I've run into a new problem.

Integral of: x^5cos(x^3)

I guess that I could set x^5 = u and do it 5 times but it seems like there should be a much easier way.

Again, sorry about the typo and thanks for any help.

7. Try $u = x^3$.

Then it will break $x^5$ into $x^3 \cdot x^2$.

And the remaining $x^2$ will get eliminated when you derivate u.

8. Originally Posted by Smilie
Ah, sorry I typed the wrong integral X_X. Anyways I was able to get the answer and I've run into a new problem.

Integral of: x^5cos(x^3)

I guess that I could set x^5 = u and do it 5 times but it seems like there should be a much easier way.

Again, sorry about the typo and thanks for any help.
Have you gone to this site to see what the antiderivative might be?
The Integrator--Integrals from Mathematica

9. ## ...

Quote:
Originally Posted by Smilie
Ah, sorry I typed the wrong integral X_X. Anyways I was able to get the answer and I've run into a new problem.

Integral of: x^5cos(x^3)

I guess that I could set x^5 = u and do it 5 times but it seems like there should be a much easier way.

Again, sorry about the typo and thanks for any help.

Have you gone to this site to see what the antiderivative might be?
The Integrator--Integrals from Mathematica

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Actually I have the answer to the integral, I'm just not sure how to solve it.

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Try u=x^3.

Then it will break into .

And the remaining will get eliminated when you derivate u.

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Unfortunately I still have no idea what to do.

So x^3 would be u, du = 3x^2dx
I"m not quite sure how that cancels the x^2 or whatever. Anyway someone can show how to work the problem? As I said, I already have the answer I just need to see how to solve it.

Thanks.

10. $\int x^5 \cos (x^3) dx = \int x^2 x^3 \cos (x^3) dx \ \ \ \ \ [ u = x^3 \longrightarrow du = 3x^2dx ]$

$= \int \frac{u x^2 cos(u)}{3x^2} du = \tfrac{1}{3} \int u \underbrace{cos(u)}_{\tfrac{d}{du} \sin (u)} du = \tfrac{1}{3} ( u \sin (u) - \int \sin u du )$

$= \tfrac{1}{3} [ u \sin (u) + cos (u)] = \tfrac{1}{3} [ x^3 \sin (x^3) + cos (x^3)]$

(I hope I didn't just do your homework for you.)

You really need to understand what I did there. It's for you're own benefit.

11. ## Thanks!

Thanks Trevor, I got it. I was able to get most of the problem but for some reason when I substituted for U i didn't change the x^3 in the cos to U and was wondering how I was supposed to integrate that.

Also don't worry about doing my homework, these problems were all done in a group a week ago, I'm just going back and figuring out how to do them X_X.

Thanks.

12. Next time, when you need to integrate $f(x)g(x)$ and both functions are not cyclical when derived (i.e. not exponential and trigonometric), then you can use tabular integration with ease.