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karisos

  • 3 years ago

Evaluate the integral by using multiple substitutions ∫5(3x^2-7)sin^5(x^3-7x) cos(x^3-7x) dx

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  1. karisos
    • 3 years ago
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    \[\int\limits 5(3x^2-7)\sin^5(x^3-7x)\cos(x^3-7x)dx\]

  2. Algebraic!
    • 3 years ago
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    u= x^3−7x du = 3x^2-7 dx

  3. Lilith
    • 3 years ago
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    well \[u=x^3-7x;du=3x^2-7\] this leaves \[\int5sin^5(u)cos(u)du\]

  4. Algebraic!
    • 3 years ago
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    v =sin (u) dv = cos(u) du

  5. Lilith
    • 3 years ago
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    so the answer would be \[5sin^6(x^3-7x)\]

  6. karisos
    • 3 years ago
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    Makes sense... THANK YOU BOTH!! ^_^

  7. karisos
    • 3 years ago
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    T.T i thought i could give both of u medals T.T

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