Question

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

Answer 1

crud I just realized where I went wrong

Answer 2

i mean antiderivative of a product

Answer 3

do I at least have the right idea with the substituion rule or am I way off?

Answer 4

∫ 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!

Answer 5

man this migrane isn't helping me integrate integration by parts into my head