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Math Help - integrals

  1. #1
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    integrals

    I can not figure out how to do this question. Do I seperate it into its sides first or something? Do I treat the nuber in front with the x as a constant? Im confused. Please show me what to do.

    integrate (7x^3 cos(2+x^4) - 8x^3 e^(2+x^4))dx
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  2. #2
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    Re: integrals

    Hello, Twinkie!

    I cannot figure out how to do this question.
    Do I seperate it into its sides first or something? . What does that mean?
    Do I treat the number in front with the x as a constant? . Of course!
    I'm confused. . So am I.
    Please show me what to do.

    \int \left[7x^3\cos(x^4+2) - 8x^3e^{x^4+2}\right]dx

    We have two integrals: . \begin{Bmatrix}7\int x^3\cos(x^4+2)\,dx & [1] \\ \\ -8\int x^3e^{x^4+2}\,dx & [2] \end{Bmatrix}



    [1]\;\text{Let }u \,=\,x^4+2 \quad\Rightarrow\quad du \,=\, 4x^3\,dx \quad\Rightarrow\quad x^3\,dx \,=\,\tfrac{1}{4}du

    Substitute: . 7\int\cos(x^4+2)\left(x^3\,dx\right) \;=\;7\int\cos u\left(\tfrac{1}{4}\,du\right)
    . . . . . . . . =\;\tfrac{7}{4}\int \cos u\,du \;=\;\tfrac{7}{4}\sin u + C_1

    Back-substitute: . \tfrac{7}{4}\sin(x^4+2) + C_1



    [2]\;\text{Let }u \,=\,x^4+2 \quad\Rightarrow\quad du \,=\, 4x^3\,dx \quad\Rightarrow\quad x^3\,dx \,=\,\tfrac{1}{4}du

    Substitute: . -8\int e^{x^4+2}\left(x^3\,dx\right) \;=\;-8\int e^u\left(\tfrac{1}{4}\,du\right)
    . . . . . . . . =\;-2\int e^u\,du \;=\;-2e^u + C_2

    Back-substitute: . -2e^{x^4+2} + C_2



    Answer: . \tfrac{7}{4}\sin(x^4+2) - 2e^{x^4+2} + C
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  3. #3
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    Re: integrals

    What part of the equaision did the X^3 dx=(1/4)du come from?
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  4. #4
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    Re: integrals

    Quote Originally Posted by Twinkie View Post
    What part of the equaision did the X^3 dx=(1/4)du come from?

    Here let:

    u = x^4 + 2

    If you take the derivative of the above assumption with respect to x you'll get:

    \frac{\mathrm{d}u}{\mathrm{d}x} = 4x^3

    Then multiply both sides of this by \tfrac{1}{4}\,\mathrm{d}x and you'll get:

    \frac{1}{4}\,\,\mathrm{d}u = x^3\,\,\mathrm{d}x

    And hence:

    x^3\,\,\mathrm{d}x = \frac{1}{4}\,\,\mathrm{d}u

    Hope it answers your question.
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  5. #5
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    Re: integrals

    Yes, thank you!
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