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cuddlepony
 3 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
cuddlepony
 3 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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cuddlepony
 3 years ago
Best ResponseYou've already chosen the best response.0crud I just realized where I went wrong

cuddlepony
 3 years ago
Best ResponseYou've already chosen the best response.0i mean antiderivative of a product

cuddlepony
 3 years ago
Best ResponseYou've already chosen the best response.0do I at least have the right idea with the substituion rule or am I way off?

ezhyl_marie
 3 years ago
Best ResponseYou've already chosen the best response.1∫ 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!

cuddlepony
 3 years ago
Best ResponseYou've already chosen the best response.0man this migrane isn't helping me integrate integration by parts into my head
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