Integration by Parts

[Austic]

Ok so my problem asks me to take the integral of (x^5)cos(x^3)dx by first doing u-substitution (or in my case y-substitution to make everything more organized) then integrate by parts.

So y=x^3 and dy=2y^2 (or 1/2dy=y^2), then you set your integral up as 1/2(integral of)ycos(y)dy, then from there I integrated by parts

I set u=y and dv=cosy
du=1dy v=siny

Then I plugged everything in and got 1/2[ysiny-(integral of)siny]
1/2[ysiny + cosy]
1/2ysiny + 1/2cosy

Then substitute the original x^3 back in for y and get 1/2(x^3)sin(x^3) + 1/2cos(x^3).

I did all the work and my online web homework is telling me that it is wrong. What did I mess up on?

 

[galactus]

You are on the right track, but have a small error.


Note that \(\displaystyle y=x^{3}, \;\ dy=3x^{2}dx, \;\ \frac{dy}{3}=x^{2}dx\)

\(\displaystyle \int x^{3}\cdot x^{2}cos(x^{3})dx\)

\(\displaystyle \frac{1}{3}\int ycos(y)dy\)

Let \(\displaystyle u=y, \;\ dv=cos(y), \;\ du=dy, \;\ v=sin(y)\)

\(\displaystyle \frac{1}{3}\left[ysin(y)-\int sin(y)\right]dy\)

\(\displaystyle \frac{1}{3}\left[x^{3}sin(x^{3})+cos(x^{3})\right]\)

 

[Austic]

Wow, I feel like such an idiot

Sorry, I've been doing these kind of problems all day and I guess I'm just so burnt out I didn't even realize my simple mistakes