This can be done by using the first part of the fundamental thereom of calculus,

The first part of the theorem says,if,

`F(x) = int_a^xf(t)dt ` then,

`F'(x) = f(x) `

Now our integral is,

`F(x) = int_cos(x)^sin(x)log(5+9v)dv`

`F(x) = int_cos(x)^0log(5+9v) dv + int_0^sin(x)log(5+9v)dv`

`F(x) = -int_0^cos(x)log(5+9v) dv + int_0^sin(x)log(5+9v)dv`

now let us separate this into two integrals,

such that,

`F(x) = y + z`

where,

`y = -int_0^cos(x)log(5+9v)dv` and

`z = int_0^sin(x)log(5+9v)dv`

Now we can evaluate the derivatives of two functions separately,

first, y

we shall make a substitution as u =cos(x)

then `(du)/(dx) = -sin(x)`

so y becomes,

`y =-int_0^ulog(5+9v)dv`

now from the theroem, you get,

`(dy)/(du) = log(5+9u)`

Now from the chain rule, you get,

`(dy)/(dx) = (dy)/(du)*((du)/(dx)`

`(dy)/(dx) = log(5+9u)*(-sin(x))`

`(dy)/(dx) = -sin(x)log(5+9cos(x))`

Now we can floow the same procedure for z also,

let t = sin(x)

then `(dt)/(dx) = cos(x)`

now z becomes,

`z =int_0^tlog(5+9v)dv`

`(dz)/(dt) = log(5+9t)`

now from the chain rule we know,

`(dz)/(dx) = (dz)/(dt) * (dt)/(dx)`

`(dz)/(dx) = log(5+9t)*cos(x)`

`(dz)/(dx) = cos(x)log(5+9sin(x))`

Now F =y +z

then

`(dF)/(dx) = (dy)/(dx) + (dz)/(dx)`

`(dF)/(dx) = -sin(x)log(5+9cos(x)) + cos(x)log(5+9sin(x))`

`(dF)/(dx) = cos(x)log(5+9sin(x)) - sin(x)log(5+9cos(x))`

`(dF)/(dx) = log(5+9sin(x))^cos(x) - log(5+9cos(x))^sin(x)`

`(dF)/(dx) = log[((5+9sin(x))^cos(x))/((5+9cos(x))^sin(x))]`