Using the substitution x = 2cosu, find the integral of dx/((x^2)(4-x^2)^1/2), evaluated between x=1 and x=sqrt(2).

Starting with x=2cosu, rearrange for u to get u=arccos(x/2), then find the upper and lower limits of the integral. We find that our lower limit goes from 1 to pi/3, ad our upper limit goes from root 2 to pi/4.
Now to find the derivative of x, dx, in terms of the derivative of u, du. Find dx/du, which is simply
dx/du = -2sinu.
Thus, dx = -2sinudu.
Now substitute our x's for our equations in terms of u, as well as the upper and lower limits. The root(4-x²) term becomes root(4-4cos²u), which simply becomes 2sinu after we make use of the identity sin²x = 1 - cos²x. This 2sinu term in the denominator cancels with the 2sinu term in the numerator, which we got from subbing dx with -2sinudu.
Now our integral is simply -1/4 sec²u du, giving us
I = -1/4 tanu,
evaluated between u= pi/4 and u = pi/3. We may flip our limits and multiply our equation by minus 1 so that our upper limit is now a larger number than the lower limit, but this does not change the answer.
Evaluate to get the answer to be (-1+sqrt(3))/4.