- #1

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## Homework Statement

[tex]\int x^5cos(x^3) dx[/tex]

## Homework Equations

[tex]\int uv' = uv - \int u'v[/tex]

## The Attempt at a Solution

[tex]\int x^5cos(x^3) dx[/tex]

[tex]\frac{(x^5)*(sin(x^3)}{(3x^2)} - \int\frac{5x^4*sin(x^3)}{(3x^2)} dx[/tex]u = [tex]x^5[/tex]

du = [tex]5x^4[/tex]

v = [tex] \frac{sin(x^3)}{(3x^2)}[/tex]

dv = [tex]cos(x^3)[/tex]

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

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

[tex]\int x^2*sin(u) *\frac{du}{3x^2}[/tex]u = [tex]x^3[/tex]

du = [tex]3x^2 dx[/tex]

dx = [tex] \frac{du}{3x^2}[/tex]

[tex]\frac{1}{3} \int sin(u) du[/tex]

[tex]\frac{-1}{3} cos(u)[/tex]

[tex]\frac{-1}{3} cos(x^3)[/tex]

Revisting

[tex]\frac{(x^3)*sin(x^3)}{3} - \frac{5}{3} * \frac{-1}{3} cos(x^3)[/tex]

**My answer:**

[tex]\frac{(x^3)*sin(x^3)}{3} + \frac{5}{9} * cos(x^3) + C[/tex]

**Correct Answer:**

[tex]\frac{(x^3)*sin(x^3)}{3} + \frac{1}{3} * cos(x^3) + C[/tex]