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

Staff Emeritus
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