Asymptotic behavior of $\int_0^1 \text{min}\{1,x^{-d}(1-x^d)^n\} dx$ as $n\to\infty$ I am trying to find bounds on 
\begin{equation*}
\int_0^1 \text{min}\{1,x^{-d}(1-x^d)^n\} dx~\text{as}~n\to\infty,~d\in\mathbb{N}. 
\end{equation*}
I found this integral trying to bound some probability, therefore, I have no hope for a closed form and I am only interested in the behavior as $n\to\infty$, for fixed $d\in\mathbb{N}$.
Unfortunately, I can't even get rid of the minimum myself.

In order to estimate our integral, we just need to find in which point $x\in(0,1)$ we have:
$$ (1-x)^n = x. \tag{1}$$
By exploting convexity, we may use Newton's method with starting point in the origin to prove that the solution of $(1)$ is greater than $\frac{1}{n+1}$. That gives that our integral is trivially $\geq\frac{1}{(n+1)^{1/d}}$.
On the other hand,
$$ \int_{0}^{1}\min\left(1,x^{-d}(1-x^d)^n\right)\,dx\leq\frac{1}{(n+1)^{1/d}}+\int_{(n+1)^{-1/d}}^{1}x^{-d}(1-x^d)^n\,dx\tag{2} $$
and:
$$\begin{eqnarray*}\int_{(n+1)^{-1/d}}^{1}x^{-d}(1-x^d)^n\,dx&=&\frac{1}{d}\int_{\frac{1}{n+1}}^{1}y^{1/d-2}(1-y)^n\,dy\\&\leq&\frac{1}{d}\int_{\frac{1}{n+1}}^{+\infty}y^{1/d-2}e^{-ny}\,dy\\&\leq&\frac{e\,(n+1)^{1/d}}{(3d-1)(n+1)^3}\tag{3}\end{eqnarray*}$$
by estimating the last integral with minus the inverse of the derivative of the integrand function in the left endpoint. So we have that the original integral is not very far from being $\frac{1}{(n+1)^{1/d}}$.
