Evaluate $\lim_{n\to\infty}\int_0^{\frac {\pi}{3}}\frac {\sin^nx}{\sin^nx+\cos^nx}dx$
I tried using the substitution $u=\frac {\pi}{2}-x$ and maybe thought this function might be symmetrical in some way and got: $$\lim_{n\to\infty}\int_{\frac {\pi}{6}}^{\frac {\pi}{2}}\frac {\cos^nx}{\sin^nx+\cos^nx}dx$$
but it's not really helping me... any other ideas?
A heuristic solution. Interchanging the order of integral and limit, we have
$$ \lim_{n\to\infty} \int_{0}^{\frac{\pi}{3}} \frac{\sin^n x}{\sin^n x+\cos^n x} \, dx = \int_{0}^{\frac{\pi}{3}} \lim_{n\to\infty} \frac{\sin^n x}{\sin^n x+\cos^n x} \, dx. $$
Now depending on the relative size of $\sin x$ and $\cos x$, we have
$$ \lim_{n\to\infty}\frac{\sin^n x}{\sin^n x+\cos^n x} = \begin{cases} 0, & \text{if } 0 \leq x < \frac{\pi}{4} \\ \frac{1}{2}, & \text{if } x = \frac{\pi}{4} \\ 1, & \text{if } \frac{\pi}{4} < x \leq \frac{\pi}{3}. \end{cases} \tag{1} $$
So it follows that the limit is $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} dx = \frac{\pi}{12}$.
Remark. Indeed we took a huge leap by assuming that the integral and the limit can be switched. This is indeed possible in our case, although a direct justification required some advance results such as dominated convergence theorem.
A preliminary analysis level solution. Fix any sufficiently small $\epsilon > 0$ and consider $a=\frac{\pi}{4}-\epsilon$ and $b = \frac{\pi}{4}+\epsilon$. If we write $I_n = \int_{0}^{\frac{\pi}{3}} \frac{\sin^n x}{\sin^n x+\cos^n x} \, dx$, then the integrand is monotone increasing in $x$ and hence
\begin{align*} I_n &\geq \int_{b}^{\frac{\pi}{3}} \frac{\sin^n x}{\sin^n x+\cos^n x} \, dx \\ &\geq \int_{b}^{\frac{\pi}{3}} \frac{\sin^n b}{\sin^n b+\cos^n b} \, dx \\ &= \left(\frac{\pi}{3}-b\right)\frac{\sin^n b}{\sin^n b+\cos^n b} \end{align*}
This gives
$$ \liminf_{n\to\infty} I_n \geq \lim_{n\to\infty} \left(\frac{\pi}{3}-b\right)\frac{\sin^n b}{\sin^n b+\cos^n b} = \frac{\pi}{12}-\epsilon.$$
But since the LHS of the above inequality is a constant independent of $\epsilon$, letting $\epsilon \downarrow 0$ proves that $\liminf_{n\to\infty} I_n \geq \frac{\pi}{12}$. Similarly,
\begin{align*} I_n &\leq \int_{0}^{a} \frac{\sin^n x}{\sin^n x+\cos^n x} \, dx + \int_{a}^{\frac{\pi}{3}} \frac{\sin^n x}{\sin^n x+\cos^n x} \, dx \\ &\geq \int_{0}^{a} \frac{\sin^n a}{\sin^n a+\cos^n a} \, dx + \int_{a}^{\frac{\pi}{3}} dx \\ &= a\frac{\sin^n a}{\sin^n b+\cos^n a} + \left(\frac{\pi}{3}-a\right) \end{align*}
and hence
$$ \limsup_{n\to\infty} I_n \leq \lim_{n\to\infty} \left[ a\frac{\sin^n a}{\sin^n b+\cos^n a} + \left(\frac{\pi}{3}-a\right) \right] = \frac{\pi}{12}+\epsilon. $$
Letting $\epsilon \downarrow 0$, we obtain $\limsup_{n\to\infty} I_n \leq \frac{\pi}{12}$. These together tell that
$$\liminf_{n\to\infty} I_n = \limsup_{n\to\infty} I_n = \frac{\pi}{12}$$
and therefore the limit of $I_n$ exists and has the value $\frac{\pi}{12}$.
Addendum. The following demonstrates graphs of the integrand for different values of $n$.
$\hspace{5em}$
The graph is already quite close to $\text{(1)}$ when $n=100$, which provides a sanity check.