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 $x$.
 A: 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
$$\frac{\pi}{2} \leq \int_0^1 \frac{dx}{\sqrt{1-x^2 - \epsilon(1-x^3)}} \leq \frac{\pi}{2} + \frac{3\pi}{8} \epsilon + O(\epsilon^2)$$
A: 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...
$$I(\epsilon) = \frac{\pi}{2} + \epsilon + O(\epsilon^2)$$
A: 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 $\epsilon \to 0$
$$-\arctan\left(\frac{1}{\sqrt{-\epsilon}}\right) = \frac{\pi}{2}$$
