Simple Integration problem: Trigonometric Integration

I'm not sure what I did wrong here, or if the integral calculator has given me an equivalent answer, I'll try my best to explain.

Question: find the indefinite integral of cos(x)*cos^5(sin(x))dx

First step is easy, to sub u=sin(x), du=cos(x)dx

= int cos^5(u)du

Then I subbed the Cos^2(x)=1-Sin^2(x) formula into the equation to get

= int (1-sin^2(u))^2*cos(u)du then sub v=sin(u), dv=cos(u)du
= int (1-v^2)^2*dv
= int v^4 - 2v^2 + 1
= 1/5*v^5 - 2/3*v^3 + v + C
= 1/5*sin^5(u) - 2/3*sin^3(u) + sin(u) + C
= 1/5*sin^5(sin(x)) - 2/3*sin^3(sin(x)) + sin(sin(x)) + C

An integral calculator gives an answer of 1/80*(sin(5*sin(x)) + 5/48(sin(3*sin(x)) + 5/8*(sin(sin(x))

Thats alot different from my answer, what did I do wrong?

Hello.
Am sure, your solution is correct.
Maybe, the calculator's solution is a different form for your answer as you said.

Hi

You did nothing wrong since

1/5*sin^5(sin(x)) - 2/3*sin^3(sin(x)) + sin(sin(x)) = 1/80*(sin(5*sin(x)) + 5/48(sin(3*sin(x)) + 5/8*(sin(sin(x))