$\lim_{n\to\infty}\int_0^{10}(1-|\sin t|)^n nd t=1$? $\lim_{n\to\infty}\int_0^{10}(1-|\sin t|)^n nd t=1$?
I use numerical experients, and showed that the above limit is valid. But what is the analytical proof?
It sounds that integrating by parts is not sufficient.
 A: By symmetry we can rewrite the integral as
$$\lim_{n\to\infty}7\int_0^{\frac{\pi}{2}}n(1-\sin t)^n\:dt + \int_{10}^{\frac{7\pi}{2}} n(1+\sin t)^n\:dt$$
The term on the right goes to $0$ by squeeze theorem since we have that
$$0\leq \int_{10}^{\frac{7\pi}{2}} n(1+\sin t)^n\:dt \leq n(1+\sin 10)^n\cdot\left(\frac{7\pi}{2}-10\right) \to 0$$
For the remaining piece use the substitution $x = (1-\sin t)^{n+1}$
$$|dx| = (n+1)\cdot(1-\sin t)^n\cdot \cos t\;dt $$
$$\implies \lim_{n\to\infty}\frac{7n}{n+1}\int_0^1 \frac{dx}{\sqrt{1-\left(1-x^{\frac{1}{n+1}}\right)^2}} \longrightarrow 7\int_0^1 dx = 7$$
by dominated convergence.
A: This is a typical example of the approximation to the identity. The relevant result here is:

Claim. Let $a \in (0, \frac{\pi}{2}]$ and $f$ be any bounded measurable function on $[0, a]$ that is continuoust at $x = 0$. Then
$$ \lim_{n\to\infty} \int_{0}^{a} f(x) \cdot n (1 - \sin x)^n \, \mathrm{d}x = f(0). $$

Once we have this result, then using that $3\pi < 10 < \frac{7\pi}{2}$ and the symmetry of $\sin$,
\begin{align*}
\int_{0}^{10} n (1-\left|\sin x\right|)^n \, \mathrm{d}x
&= 6 \biggl( \int_{0}^{\frac{\pi}{2}} n (1-\sin x)^n \, \mathrm{d}x \biggr) + \int_{0}^{\frac{7\pi}{2}-10} n (1-\sin x)^n \, \mathrm{d}x.
\end{align*}
By the claim, this converges to $6 \cdot 1 + 1 = 7$ as $n\to\infty$.

Proof of Claim. Substituting $x = s/n$,
$$ \int_{0}^{a} f(x) \cdot n (1 - \sin x)^n \, \mathrm{d}x = \int_{0}^{\infty} f(s/n) \left(1 - \sin(s/n)\right)^n \mathbf{1}_{[0,na]}(s) \, \mathrm{d}s $$
On the other hand, the concavity of $\sin$ on $[0, \frac{\pi}{2}]$ shows that $\sin x \geq cx$ on this interval with $c = \frac{2}{\pi}$. Then, from this and the inequality $1 - t \leq e^{-t}$ together,
$$ (1 - \sin(s/x))^n \leq e^{-n \sin(s/x)} \leq e^{-cs}. $$
This shows that $f(s/n) \left(1 - \sin(s/n)\right)^n \mathbf{1}_{[0,na]}(s)$ is dominated by an integrable function $(\sup |f|) e^{-cs}$, and so, the dominated convergence theorem shows that
\begin{align*}
\lim_{n\to\infty} \int_{0}^{a} f(x) \cdot n (1 - \sin x)^n \, \mathrm{d}x
&= \int_{0}^{\infty} \lim_{n\to\infty}  f(s/n) \left(1 - \sin(s/n)\right)^n \mathbf{1}_{[0,na]}(s) \, \mathrm{d}s \\
&= \int_{0}^{\infty} f(0) e^{-s} \, \mathrm{d}s \\
&= f(0).
\end{align*}
