Consider the function ${I}\left(x\right)$ defined by the integral$${I}\left(x\right) =\int_{x}^{\pi}\left[\frac{1 - \cos\left(y\right)}{\cos\left(x\right) - \cos\left(y\right)}\right]^{1/2}{\rm d}y,\qquad x \in \left[0,\pi\right]$$
- The integral can be calculated exactly using a series of variable substitutions ( e.g. one can set $y = \pi - 2\alpha$, use the double angle formula for the cosine function and then set $u = \sin\left(\alpha\right)$ to calculate it ), and the result is that ${I}\left(x\right) = \pi$ independent of $x$.
- Another way to show this result would be to note that $I(0) = \pi$ ( since for $x = 0$ the integrand is just $1$ ) and then show that the derivative of $I$ is zero everywhere.
- The latter can of course be done using the same substitutions as mentioned above.
Question. But I would like to know if it is possible to show that ${I}\left(x + \epsilon\right) = {I}\left(x\right) + o\left(\epsilon\right)$ directly without the substitutions ? (Of course we actually have ${I}\left(x+\epsilon\right) = {I}\left(x\right)$, but suppose we don't know that.)