<s>[INST] 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 $x$.
 [/INST] You can be authorized in doing that because $\epsilon$ is small (we suppose small enough to give you the "license" of dong that). 
Also notice that the range of the integral is $[0, 1]$, hence $x$ too, in a certain way, is small, and terms like $x^3$ are then even smaller.
There are many approaches for dealing with such an integral. Yours is one. Then you can take that function and make a Tylor series in $x$ or in $\epsilon$, obtaining respectively:
$$\star : \frac{1}{\sqrt{1-\epsilon }}+\frac{x^2}{2 (1-\epsilon )^{3/2}}-\frac{x^3 \epsilon }{2 (1-\epsilon )^{3/2}}+O\left(x^4\right)$$
$$\star : \frac{1}{\sqrt{1-x^2}}-\frac{\left(x^3-1\right) \epsilon }{2 \left(1-x^2\right)^{3/2}}+\frac{3 \left(x^3-1\right)^2 \epsilon ^2}{8 \left(1-x^2\right)^{5/2}}-\frac{5 \left(x^3-1\right)^3 \epsilon ^3}{16 \left(1-x^2\right)^{7/2}}+O\left(\epsilon ^4\right)$$
Those are clearly Taylor series of the whole function, that is, of
$$\frac{1}{\sqrt{1 - \epsilon - x^2 + \epsilon x^3}}$$
Another way is to think about the term $\epsilon x^3$ and rawly say "let's get rid of this", remaining with the easy integral
$$\int_0^1 \frac{\text{d}x}{\sqrt{1 - \epsilon - x^2}} = -\tan ^{-1}\left(\frac{\epsilon }{(-\epsilon )^{3/2}}\right)$$
Numerica other methods are available, but this wouldn't be the topic of the question.
EDIT
Notice that in my last result we have
$$\frac{\epsilon}{(-\epsilon)^{3/2}} = \frac{1}{\sqrt{-\epsilon}}$$
hence as $\epsilon \to 0$
$$-\arctan\left(\frac{1}{\sqrt{-\epsilon}}\right) = \frac{\pi}{2}$$
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