G5wediannaMarie
contestada

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

Respuesta :

caylus
Hello,

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


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