Let g(x) = ∫ from cosx to 1 ( 1−𝑡^2 / 1+𝑡 ^4 )𝑑𝑡 . Determine 𝑔′(𝑥).

Answer 1

We use the Fundamental Theorem of Calculus. cos(x) is on the lower end of the integral, but there is no x-term on the upper end. As it is on the bottom end, we take the negative of it. Thus, -(1-t^2)/(1+t^4) evaluated at cos(x) is -(1- cos²(x))/(1+cos⁴(x)) = -sin²(x)/(1+cos⁴(x))