1. Feb 6, 2008

### mckallin

can anyone help me solve the following integration? thanks a lot.

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

2. Feb 6, 2008

### transgalactic

try to solve it by parts

3. Feb 6, 2008

### mckallin

I have tried it, but it doesn't work, at least, for me.

If I make $$u=cos(x^3), dv=x^5 dx$$, the grade of x, which is in the $$cos(x^3)$$, won't be reduce.

If I make $$u=x^5, dv=cos(x^3) dx$$, I can't solve the $$\int cos(x^3) dx$$.

Could you give me some more advice? thanks

4. Feb 6, 2008

### Tedjn

What would you like to have inside the integral in order to have

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

be solvable? Can you choose slightly different u and dv to accomplish that?

5. Feb 6, 2008

### Rainbow Child

From $(\sin x^3)'=3\,x^2\,\cos x^3$ deduce that $\cos x^3=\frac{1}{3\,x^2}\,(\sin x^3)'$ and use that to integrate by parts.

6. Feb 6, 2008

### HallsofIvy

Oh, for goodness sake! Make the substitution u= x3, then integrate by parts!

(I notice now that that is essentially what Rainbow Child said.)

7. Feb 6, 2008

### fermio

How it can help?

How this can help?

8. Feb 6, 2008

### Hootenanny

Staff Emeritus
I don't think there's any harm in showing the substitution,

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

Hence when we make the substitution the integral becomes,

$$\int \frac{x^5}{3x^2}\cos(u)du = \frac{1}{3}\int u\cos(u)du$$

Which is a simple integral to solve.

Last edited: Feb 6, 2008