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

Hello,


`\int{x*cos(5x)} \, dx=x*(sin(5x))/(5)- (1)/(5)*\int{sin 5x} \, dx\\=(x*sin(5x))/(5)+(cos(5x))/(25)+c`

Answer 2

I think that you have wrong u-substitution ( partial integration ):
`\int {u} \, dv= uv- \int {v} \, du`
u=x, dv=cos 5x dx
du=dx, v=1/5 * sin 5x
Integral becomes:
`= (xsin 5x)/(5)- \int { (1)/(5) sin 5x} \, dx= \\ = (xsin5x)/(5)- (1)/(5) (- (1)/(5)cos5x)= (x sin5x)/(5)+ (cos5x)/(25) +C`
Is it OK now?