# Math Help - Integration discrepancy with wolfram Online

1. ## Integration discrepancy with wolfram Online

Integrate:

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

This is what I did:
Let $u=sin(x)+4$

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

$du=cos(x)dx$
Therefore, I have:

$\int_ (u^3)du$

Now,I get

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

Subbing back in, I get

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

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

Then, quartering everything:

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

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

Where have I gone wrong??
2. 64 is a constant, so the answer remains the same whether you have the 64 or you don't have it. Wolfram's answer is different than yours in that they have the constant of integration $c' = c + 64$ where your constant is c.