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Thread: Integration by Parts

  1. #1
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    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.
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    Quote Originally Posted by Smilie View Post
    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 $\displaystyle u = 3x$. Then do your integration by parts.

    -Dan
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  3. #3
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    Quote Originally Posted by Smilie View Post
    Hello,


    The Integral of: 27x^2/cos(3x)
    Maybe this is a "wrong" integral?
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  4. #4
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    Quote Originally Posted by curvature View Post
    Maybe this is a "wrong" integral?
    Certainly looks that way:
    Attached Thumbnails Attached Thumbnails Integration by Parts-msp2.gif  
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  5. #5
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    Maybe the integrand is this: $\displaystyle \frac{27x^2}{cos(x^3)}$. That makes the substitution much easier.
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    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.
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  7. #7
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    Try $\displaystyle u = x^3$.

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

    And the remaining $\displaystyle x^2$ will get eliminated when you derivate u.
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  8. #8
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    Quote Originally Posted by Smilie View Post
    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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  9. #9
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    ...

    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

    -----------------------------------------------------------------

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

    -----------------------------------------------------------------

    Try u=x^3.

    Then it will break into .

    And the remaining will get eliminated when you derivate u.

    -----------------------------------------------

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

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

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


    (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.
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  11. #11
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    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.
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  12. #12
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    Next time, when you need to integrate $\displaystyle f(x)g(x)$ and both functions are not cyclical when derived (i.e. not exponential and trigonometric), then you can use tabular integration with ease.
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