Find the derivative of

Integral from -5 to sin(x) of cos(t^3)+t

To find the derivative of the integral, you can use the Fundamental Theorem of Calculus. According to the theorem, if you have an integral of the form

∫f(g(x))g'(x) dx,

where g(x) is the upper limit of the integral, then the derivative of the integral is simply f(g(x)).

In your case, you have

∫[-5 to sin(x)] (cos(t^3) + t) dt.

To find its derivative, we need to identify f(g(x)) and g'(x) in the fundamental theorem formula.

Let f(t) = cos(t^3) + t. The integral is taken with respect to t, so f(t) is the function being integrated.

Let g(x) = sin(x). The upper limit of the integral is g(x).

Now, we need to find the derivative of the integral. According to the Fundamental Theorem of Calculus, the derivative will be f(g(x)), which means we need to substitute g(x) into f(t).

So, substitute g(x) = sin(x) into f(t) = cos(t^3) + t:

f(g(x)) = cos((sin(x))^3) + sin(x).

Therefore, the derivative of the given integral is:

d/dx ∫[-5 to sin(x)] (cos(t^3) + t) dt = cos((sin(x))^3) + sin(x).