find d/dc intergral[-c,a] (sin(s^3t))ds

[kpx001]

using the fundamental of calculus rule i swap a with b and get
-intergral sin(s^3t)
then i plug in
-sin((-c)^3t)
sin((c^3t))
but the answer is
-sin(c^3t).
what did i do wrong?

 

[royhaas]

I think you forgot the chain rule at the end. You should have \(\displaystyle - \sin((-c)^3t)\frac {d(-c)}{dc}\).