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

### mjsd

you really should show some more effort. ie. show us what you have tried, what you know and tell us what you have problems with.

hint: try look for a substitution to make the integrand look nicer, then you can try by parts.

3. Feb 6, 2008

### mckallin

I have tried to do it by part.

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

I have thought if there is some way to make it look nicer (like $$u=x^3$$ ),but I still can't work out a better substitution.

Could you give me some more advice?

4. Feb 6, 2008

### hanskuo

try
$$u=x^3,dv=x^2\cos(x^3)dx$$

have you learned partial substitution ?

5. Feb 6, 2008

### mjsd

here I suggested a two-step process,
substitiion: u = x^3 seems ok
then by parts in new variable

then put answer back in x.