find the derivative using fundamental theorem of calculus integral of sin^3tdt from e^x to 0

Answer 1

d/dx ∫[e^x,0] sin^3(t) dt
= d/dx -∫[0,e^x] sin^3(t) dt
= -sin^3(e^x) * d/dx (e^x)
= -e^x sin^3(e^x)