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

1. Apr 30, 2009

### fallen186

1. The problem statement, all variables and given/known data

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

2. Relevant equations
$$\int uv' = uv - \int u'v$$

3. The attempt at a solution
$$\int x^5cos(x^3) dx$$

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

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

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

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

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

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

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

Revisting

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

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

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

2. Apr 30, 2009

### xaos

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

3. May 1, 2009

### HallsofIvy

Staff Emeritus
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 $u= x^3$.

4. May 2, 2009

### 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.

Last edited: May 2, 2009
5. May 2, 2009

### HallsofIvy

Staff Emeritus
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 $(2x^2)cos(x^3)$.