Question

Use part I of the Fundamental Theorem of Calculus to find the derivative of
h(x) = $$\underset{-<!--\; -\; -->1}{\overset{Sinx}{\int <!--\; \int \; -->}}cos{t}^2+tdt$$

Answer 1

\[\int\limits_{-1}^{Sinx} cost ^{2}+t dt\] is how the equation looks

Answer 2

My answer: -2cosx(cos(sin^2x))+sinx but I can't figure out why that is wrong.