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Math Help - ntegrate, but first make a substituition??? HELP!!!

  1. #1
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    ntegrate, but first make a substituition??? HELP!!!

    my book is asking me to "first make a substitution and then use integration by parts to evaluate the target" the integral i am working on is

    x^5 * cos(x^3)

    please help me out, I have no clue what they mean by this
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  2. #2
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    I think they want you to to substitute y = x^3. Try it and see what happens.
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    I don't understand how i am supposed to substitute it in, and what is y?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by pcgamer03 View Post
    I don't understand how i am supposed to substitute it in, and what is y?
    y is the variable you're changing the problem to... question: do you know what "integration by substitution" means?
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  5. #5
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    i do know what it means, i just can't figure you how you would to that and then integrate by parts. If i substitue u = x^3. i get du = 3x^2, or 1/(3x^2)du = dx. so now i need to integrate x^5 * cos(u) 1/(3x^2)du ???
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  6. #6
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    i do know what it means, i just can't figure you how you would to that and then integrate by parts. If i substitue u = x^3. i get du = 3x^2, or 1/(3x^2)du = dx. so now i need to integrate x^5 * cos(u) 1/(3x^2)du ???
    Excellent, you have the first step. Now you need to simplify (the numerator and denominator have a common factor) and then I think you will see the next step to complete the substitution part.

    Once you have gotten rid of all of the xs, you can integrate by parts normally.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by pcgamer03 View Post
    my book is asking me to "first make a substitution and then use integration by parts to evaluate the target" the integral i am working on is

    x^5 * cos(x^3)

    please help me out, I have no clue what they mean by this
    \int x^5 \cos \left( x^3 \right)~dx = \int x^3 \cdot x^2 \cos \left( x^3 \right)~dx

    we proceed by substitution

    Let u = x^3

    \Rightarrow du = 3x^2~dx

    \Rightarrow \frac 13 du = x^2 ~dx

    So our integral becomes:

    \frac 13 \int u \cos u~du

    now proceed with integration by parts
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