`y = int_(cos(x))^(sin(x))(ln(1 + 2v))dv` Find the derivative of the function.

You need to evaluate the the derivative of the function, hence, you need to use the part1 of fundamental theorem of calculus:

`y = int_a^b f(x)dx => (dy)/(dx) = f(x)` for `x in (a,b)`

If `f(x) = ln(1+2x)` , yields:

`(dy)/(dx) = ln(1+2x)|_(cos x)^(sin x)`

`(dy)/(dx) = ln(1+2sin x) - ln(1+2cos x)`

Using the properties of logarithms, yields:

`(dy)/(dx) = ln ((1+2sin x)/(1+2cos x))`

Hence, evaluating the derivative of the function, using the part 1...

You need to evaluate the the derivative of the function, hence, you need to use the part1 of fundamental theorem of calculus:

`y = int_a^b f(x)dx => (dy)/(dx) = f(x)` for `x in (a,b)`

If `f(x) = ln(1+2x)` , yields:

`(dy)/(dx) = ln(1+2x)|_(cos x)^(sin x)`

`(dy)/(dx) = ln(1+2sin x) - ln(1+2cos x)`

Using the properties of logarithms, yields:

`(dy)/(dx) = ln ((1+2sin x)/(1+2cos x))`

Hence, evaluating the derivative of the function, using the part 1 of the fundamental theorem of calculus, yields `(dy)/(dx) = ln ((1+2sin x)/(1+2cos x))`