Integration by Parts of x^5cos(x^3)

[fallen186]

Homework Statement

\int x^5cos(x^3) dx

Homework Equations

\int uv' = uv - \int u'v

The Attempt at a Solution

\int x^5cos(x^3) dx

u = x^5
du = 5x^4
v = \frac{sin(x^3)}{(3x^2)}
dv = cos(x^3)

\frac{(x^5)*(sin(x^3)}{(3x^2)} - \int\frac{5x^4*sin(x^3)}{(3x^2)} dx

\frac{(x^3)*sin(x^3)}{3} - \int\frac{(5x^2)*sin(x^3)}{(3)} dx

\frac{(x^3)*sin(x^3)}{3} - \frac{5}{3} \int x^2*sin(x^3) dx

u = x^3
du = 3x^2 dx
dx = \frac{du}{3x^2}

\int x^2*sin(u) *\frac{du}{3x^2}

\frac{1}{3} \int sin(u) du

\frac{-1}{3} cos(u)

\frac{-1}{3} cos(x^3)

Revisting

\frac{(x^3)*sin(x^3)}{3} - \frac{5}{3} * \frac{-1}{3} cos(x^3)

My answer:

\frac{(x^3)*sin(x^3)}{3} + \frac{5}{9} * cos(x^3) + C

Correct Answer:

\frac{(x^3)*sin(x^3)}{3} + \frac{1}{3} * cos(x^3) + C

 

[xaos]

no, your first u-dv doesn't work that way. think compostion chainrule and u-substitution.

 

[HallsofIvy]

The integral of cos(x^3) is not sin(x^3)/3x^2!

First write the integral as
\int x^3\left[x^2cos(x^3)\right]dx
and let u= x^3.

 

[fallen186]

May I ask what the derivative of sin(x^3)/3x^2 may be then?

Thanks, rewriting the function got me the right answer.

 

[HallsofIvy]

I don't know why you would ask that. It might be helpful to recognize that, by the chain rule, the derivative of sin(x^3) is (2x^2)cos(x^3).