Integrate:

$\displaystyle (4+sin(x))^3(cos(x))dx$

This is what I did:

Let $\displaystyle u=sin(x)+4$

$\displaystyle \frac{du}{dx}=cos(x)$

$\displaystyle du=cos(x)dx$

Therefore, I have:

$\displaystyle \int_ (u^3)du$

Now,I get

$\displaystyle \frac{u^4}{4}+c$

Subbing back in, I get

$\displaystyle \frac{1}{4}(sin(x)+4)^4$ which when expanded out gives me

$\displaystyle \frac{1}{4}(sin^4(x)+16sin^3(x)+96sin^2(x)+256sin( x)+256)$

Then, quartering everything:

$\displaystyle \frac{sin^4(x)}{4}+4sin^3(x)+24sin^2(x)+64sin(x)+6 4$

The problems is that Wolfram Online Integrator does not include the final term "64" in the answer.

Where have I gone wrong??

Thanks in advance