Evaluate the derivative

[kggirl]

Is this correct:
The problem says to evaluate the derivative:

d/dx (integral from 0 to cos[x] sqrt 1-t^2 dt)


I have:u = cos[x]
du/dx = sin[x]

(integral from 0 to cos[x] sqrt 1-t^2 dt) =

(integral from 0 to u 1 - t^1/2dt) du/dx=
sqrt 1-u dt * sin [x] =
sqrt 1- cos[x] * sin [x] =
(1-sinx)^1/2 cos x

 

[Gene]

Well...
d(cos(u))=-sin(u)du
But that doesn't matter 'cause cos(x) is just a limit of integration, not part of the integral.
The integral is
dy =(1-t²)^(1/2)dt

Two hints:
substitute t=sin(x)
sqrt(1-t²) is the equation of the top half of a unit circle.