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