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 $x$.

Using Taylor series you will end up having to justify evaluation at the improper bound of $1$, which will require further details that neither of the other answers have addressed. Instead, you could just note that
$$\frac{1-x^3}{1-x^2} = \frac{(1-x)(1+x+x^2)}{(1-x)(1+x)} = 1 + \frac{x^2}{1+x} \in [1,\tfrac32)$$ for $x \in [0,1)$ and hence
$$\frac{1}{\sqrt{1-x^2}}
\leq \frac{1}{\sqrt{1-x^2 - \epsilon(1-x^3)}}
=  \frac{1}{\sqrt{1-x^2}}\cdot\frac{1}{\sqrt{1 - \epsilon \frac{1-x^3}{1-x^2}}}
\leq \frac{1}{\sqrt{1-x^2}}\cdot\frac{1}{\sqrt{1 - \tfrac32\epsilon }}
$$
for all $0<\epsilon <\tfrac23$. 
So by integrating we get:
$$ \frac{\pi}{2}
\leq \int_0^1 \frac{dx}{\sqrt{1-x^2 - \epsilon(1-x^3)}}
\leq \frac{\pi}{2} \cdot \frac{1}{\sqrt{1 - \tfrac32\epsilon }}$$
We may then take the series expansion about $0$ of the right hand side, giving
$$\frac{1}{\sqrt{1-\tfrac32 \epsilon}} = 1 + \tfrac34 \epsilon + O(\epsilon^2)$$
Hence
$$\frac{\pi}{2} \leq \int_0^1 \frac{dx}{\sqrt{1-x^2 - \epsilon(1-x^3)}} \leq \frac{\pi}{2} + \frac{3\pi}{8} \epsilon + O(\epsilon^2)$$
