G5wediannaMarie
Answered

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

Verified Answer :

caylus

Hello,

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


HomertheGenius

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:\(= \frac{xsin 5x}{5}- \int { \frac{1}{5} sin 5x} \, dx= \\ = \frac{xsin5x}{5}- \frac{1}{5} (- \frac{1}{5}cos5x)= \frac{x sin5x}{5}+ \frac{cos5x}{25} +C\)
Is it OK now?

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