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cuddlepony

  • 4 years ago

Integrate xcos(5x) so I set g'(x) = x f(x) = cos(5x) Thus, I get (x^(2)cos(5x)/2) - Integral of 5xcos(5x) so I use substitution u = 5x du/5 = dx therefore, (x^(2)cos(5x)/2) + (1/5)integral of usin(u) giving me (x^(2)cos(5x)/2) + (5x^(2)cos(5x)/10) + c what am I doing wrong the text book and wolfram alpha claims I have the wrong answer

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  1. cuddlepony
    • 4 years ago
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    crud I just realized where I went wrong

  2. cuddlepony
    • 4 years ago
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    i mean antiderivative of a product

  3. cuddlepony
    • 4 years ago
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    do I at least have the right idea with the substituion rule or am I way off?

  4. ezhyl_marie
    • 4 years ago
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    ∫ x cos(5x) dx = integrate it by parts, assuming: x = u → dx = du cos(5x) dx = dv → (1/5) sin(5x) = v thus, recalling by parts integration rule, ∫ u dv = u v - ∫ v du, you get: ∫ x cos(5x) dx = (1/5)x sin(5x) - ∫ (1/5) sin(5x) dx = (1/5)x sin(5x) - (1/5) ∫ sin(5x) dx = (1/5)x sin(5x) - (1/5) [- (1/5)cos(5x)] + C = (1/5)x sin(5x) + (1/25) cos(5x) + C thus, in conclusion: ∫ x cos(5x) dx = (1/5)x sin(5x) + (1/25) cos(5x) + C I hope it helps.. Bye!

  5. cuddlepony
    • 4 years ago
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    man this migrane isn't helping me integrate integration by parts into my head

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