Evaluate ∫01cos⁡(𝑔(𝑥))⋅𝑔′(𝑥) 𝑑𝑥 given that 𝑔(0)=-5, and 𝑔(1)=

The integral ∫01cos⁡(𝑔(𝑥))⋅𝑔′(𝑥) 𝑑𝑥 can be solved using the Fundamental Theorem of Calculus.

The Fundamental Theorem of Calculus states that if a function f is continuous over the interval [a, b] and F is an antiderivative of f on [a, b], then ∫abf(x)dx = F(b) - F(a).

In this case, we can see that the integrand is the derivative of the function F(x) = sin(g(x)). This is because the derivative of sin(g(x)) with respect to x is cos(g(x)) * g'(x), which is exactly our integrand.

Therefore, we can apply the Fundamental Theorem of Calculus directly to get:

∫01cos⁡(𝑔(𝑥))⋅𝑔′(𝑥) 𝑑𝑥 = sin(g(1)) - sin(g(0))

We are given that g(0) = -5 and g(1) = ?. Unfortunately, without the value of g(1), we cannot evaluate this expression.