real analysis - Compute $lim_{nrightarrowinfty}int_{0}^{frac{pi}{2}}frac{ncos x sin^n x}{1+x}dx$

I'm trying to compute $\lim_{n\rightarrow\infty}\int_{0}^{\frac{\pi}{2}}\frac{n\cos x \sin^n x}{1+x}dx$.

In the given interval $(0,\frac{\pi}{2})$, $\frac{n\cos x \sin^n x}{1+x}$ is non negative, and increasing, I used the monotone convergence theorem to interchange the limit and the integral to get $$\lim_{n\rightarrow\infty}\int_{0}^{\frac{\pi}{2}}\frac{n\cos x \sin^n x}{1+x}dx=\int_{0}^{\frac{\pi}{2}}\lim_{n\rightarrow\infty}\frac{n\cos x \sin^n x}{1+x}dx$$

But when I'm trying to compute the limit, I'm stuck. I even tried to do through the "$u$ substitution" as well but the denominator isn't helping. Any input is appreciated.

Answer

Here's a possible approach. We have
$$
\int_0^{\pi/2} \frac{\cos x \sin^n x}{1+x } dx = \int_0^{\pi/2} \frac{\sin^n x}{1+x } d \sin x = (\text{set } y = \sin x) = \int_0^1 \frac{y^n}{ 1 + \arcsin y} dy.
$$

Now, the last integral does not seem a very pleasant one to compute explicitly, but we can compute the desired limit. Just observe that the main contribution comes from the neighborhood of 1.

Namely, fix any $0<\alpha <1$, then
$$

n \int_0^\alpha \frac{y^n}{ 1 + \arcsin y} dy \leq n \int_0^\alpha y^n dy =\frac{n}{n+1} \alpha^n \to 0 , \text{ as } n \to \infty,
$$
as $\alpha < 1$. Hence, the contribution comes from $\int_\alpha^1$, where we have
$$
\frac{y^n}{1 + \pi/2} \leq \frac{y^n}{1 + \arcsin y} \leq \frac{y^n}{1 + \arcsin \alpha}.
$$
Now integrating the last ineqalities over $[\alpha, 1]$ we obtain
$$
\frac{n}{n+1} \frac{1 - \alpha^{n+1}}{1+\pi/2} \leq n \int_{\alpha}^1 \frac{y^n}{1 + \arcsin y} dy\leq \frac{n}{n+1} \frac{1 - \alpha^{n+1}}{1 + \arcsin \alpha}.
$$

Finally, taking $n\to \infty$ in the last line, we get
$$
\frac{1}{1+\pi/2} \leq \lim n \int_{\alpha}^1 \frac{y^n}{1 + \arcsin y} dy\leq \frac{1 }{1 + \arcsin \alpha}.
$$
But we also proved that the limit of the integral over $[0, \alpha]$ is vanishing, hence the limit in the last inequality is the actual limit in question. But as $\alpha<1$ is arbitrary, we get that the limit has to be equal to
$$
\frac{1}{1+\pi/2}.
$$