How Do You Integrate x^5 cos(x^3) with Substitution?

[mckallin]

Please Help! Integration!

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

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

 

[transgalactic]

try to solve it by parts

 

[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

 

[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?

 

[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.

 

[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.)

 

[fermio]

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.

How it can help?

Make the substitution u= x³, then integrate by parts!

How this can help?

 

[Hootenanny]

How this can help?

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.