- #1

Matty R

- 83

- 0

I was hoping someone could help me with this integral.

## Homework Statement

[tex]I=\int{(x^2sin(5x^3-3))}dx[/tex]

## Homework Equations

[tex]\int{(u.\frac{dv}{dx})}dx=[uv]-\int{(v.\frac{du}{dx})}dx[/tex]

[tex]\frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dx}[/tex]

**3a. The first attempt at a solution**

[tex]u=x^2, \frac{dv}{dx}=sin(5x^3-3)[/tex]

[tex]v=-\frac{cos(5x^3-3)}{15x^2}, \frac{du}{dx}=2x[/tex]

[tex]I=[\frac{x^2.-cos(5x^3-3)}{15x^2}]-\int{((\frac{-cos(5x^3-3)}{15x^2}).2x)}dx[/tex]

[tex]So, I=-[\frac{cos(5x^3-3)}{15}]+\int{((\frac{2cos(5x^3-3)}{15x}))}dx[/tex]

**3b. The second attempt at a solution**

[tex]u=sin(5x^3-3), \frac{dv}{dx}=x^2[/tex]

[tex]v=\frac{x^3}{3}, \frac{du}{dx}=15x^2cos(5x^3-3)[/tex]

[tex]I=[sin(5x^3-3).\frac{x^3}{3}]-\int(\frac{x^3}{3}.15x^2cos(5x^3-3))[/tex]

[tex]So, I=[\frac{x^3sin(5x^3-3)}{3}]-[\frac{x^4sin(5x^3-3)}{12}][/tex]

That is as far as I've got in both situations.

I've been trying to get :

[tex]I=[something]\pm J[/tex]

[tex]J=[something]\pm I[/tex]

[tex]So, I = [something]\pm[something]\pm I[/tex]

I just can't seem to get there.

So I was wondering if someone could have a look through what I've done and see if I've gone wrong somewhere. I got through the rest of the work pretty quickly, but I'm really stuck with this one.

Thank you