$f_n(x)=n\sin^{2n+1}x\cos x$. Find $\lim_{n\to\infty}f_n(x)$

Question

Let $f_n(x)=n\sin^{2n+1}x\cos x$. Then find the value of

$\lim\limits_{n\to\infty}\displaystyle\int_0^{\pi/2}f_n(x)\;dx-\displaystyle\int_0^{\pi/2}(\lim\limits_{n\to\infty}f_n(x))\;dx$

My Thoughts: If the function is uniformly convergent, the answer should be zero.

To check for uniform convergence, I need to do Weierstrass's M-Test.

A ratio test says $f_n(x)$ is convergent provided $x<\pi/2$. How do I properly solve this problem ? Please help.

Answer 1

Here is an approach. $$ \lim\limits_{n\to\infty}\displaystyle\int_0^{\pi/2}f_n(x)\;dx = \lim\limits_{n\to\infty}n\displaystyle\int_0^{\pi/2}\sin^{2n+1}( x )\cos(x)\;dx$$

$$ =\lim\limits_{n\to\infty} \frac{n}{2n+2} \sin^{2n+2}( x )\bigg|_{x=0}^{\pi/2}=\dots\,. $$