$\displaystyle \int _c xy^4ds $ C is the right half of the circle $\displaystyle x^2+y^2=16$

Here is what i did. Since C is the right half of the circle my bounds are from $\displaystyle -\pi/2 \leq t \leq \pi/2 $. I got x = 4cost and y = 4sint. I then take the arclength of that, and multiple that by the original function with the polar coordinates substituted into it which is $\displaystyle \int (4cost)(4sint)^4\sqrt{(-4sint^2)+(4cost^2)}dt $

Im kind of confused on how to simplify the algebra before i integrate. Any step by step guidance would be appreciated

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