Taylor inside an integral I know the following integral should be: 
$$ \int_{0}^{1} \frac{dx}{\sqrt{1-x^2-\epsilon(1-x^3)}} \approx \pi/2 + \epsilon$$
for $\epsilon$ small. What I do is set $-\epsilon(1-x^3)=y$ and then since $y$ it's always greatly smaller than $(1-x^2)$ I do Taylor expansion around $y=0$:
$$ \frac{dx}{\sqrt{1-x^2+y}} \approx \frac{dx}{\sqrt{1-x^2}}-\frac{y \cdot dx}{2 \cdot (1-x^3)^{3/2}} $$ 
Then I recover $y=-\epsilon(1-x^3)$ and I get the correct answer. The problem is that I don't know if that's mathematically correct because $y$ depends on $x$.

You are on safer ground factoring the $1-x$ out of the square root as follows:
$$\begin{align} I(\epsilon) &= \int_0^1 \frac{dx}{\sqrt{1-x^2-\epsilon (1-x^3)}} \\ &= \int_0^1 dx \frac{(1-x)^{-1/2}}{\sqrt{1+x-\epsilon(1+x+x^2)}} \\ &= \int_0^1 dx \, \frac{x^{-1/2}}{\sqrt{2-x-\epsilon (3-3 x+x^2)}} \\ &=2 \int_0^1 dx \left [ 2-x^2 - \epsilon (3-3 x^2+x^4)\right ]^{-1/2} \\ &= 2 \int_0^1 dx (2-x^2)^{-1/2} \left [1-\epsilon \frac{3-3 x^2+x^4}{2-x^2} \right ]^{-1/2} \end{align}$$
I hope you see where this is heading.  You may now expand the term in brackets in a Taylor expansion in $\epsilon$ knowing that the term involving $\epsilon$ is small over the entire region of integration.  Use trig substitution and the answer is straightforwardly...
$$I(\epsilon) = \frac{\pi}{2} + \epsilon + O(\epsilon^2)$$
