A calculus problem by Naren Bhandari

Here i wish to share my solution.

We have that $2i \sin x= e^{-ix}-e^{-ix}$ with $i=\sqrt{-1}$ which follow that $P_n = \frac{1}{(2i)^{n-1}}\left(\prod_{k=1}^{n-1} e^{\frac{iki\pi}{n}}\right)\left(\prod_{k=1}^{n-1} (1-e^{\frac{ki\pi}{n}})\right)=\frac{L_{n}}{(2i)^{n-1} }\exp\left(\sum_{k=1}^{n-1} \frac{ik\pi}{n}\right)=\frac{L_n}{2^{n-1}}$ where $L_n$ is the latter product to be evaluated by noticing the polynomial $F(X)=\displaystyle \prod_{k=1}^{n}\left(X-e^{-\frac{2ki\pi}{n}}\right)$ whose zeros are non-trivial solution nth roots of unity ie $f(X)=\displaystyle\sum_{k=1}^{n-1} X^k$ and hence desired $F(1)= \displaystyle\sum_{k=1}^{n-1} 1^k=n$ and this deduce that $P_n=\displaystyle \frac{n}{2^{n-1}}$ and thus $\lim_{n\to\infty}\frac{nP_n}{2}\int_{a}^b\frac{\cos 3x}{\sin^{n}x} dx=\lim_{n\to\infty}\frac{n^2}{2^n}\int_a^b\frac{4\cos^3x -3\cos x}{\sin^n x} dx$ the latter expression can be simplified to $\displaystyle \frac{\cos x(1-4\sin^2 x)}{\sin ^n x}$ . We make u- substitution of $\sin x = u\Rightarrow \cos dx = du$ giving us $\lim_{n\to\infty}\frac{n^2}{2^n} \int_{a_1}^{b_1} \frac{1-u^2}{u^n} du=-\lim_{n\to\infty}\frac{n^2}{2^n}\frac{2^n-3n+1}{n^2-4n+3}=-1$ where $a,b$ are the lower and upper limit of the integratio so does $a_1,b_1$ after substitution.