let f(x)=x^2 and g(x)=sinx if h(x) = the integral of f(t)dt from -1 to g(x) find the value of h'(pi/6)

Question

how do i do this? not sure what to plug in to find my answer.

Answer 1

h(x) = ∫_{(from -1 to sinx)}t²dt

By the Fundamental Theorem of Calculus, h'(x) = sin²x(sinx)'

h'(x) = sin²xcosx

So, h'(π/6) = sin²(π/6)cos(π/6) = (1/2)²(√3/2) = √3/8