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Evaluate the limit

$$\displaystyle\lim_{n \to \infty} \int_{0}^{\pi} \frac{\sin x}{1+\cos ^2(nx)} dx$$

Using property of definite integral $\int_{0}^{2a} f(x).dx=2\int_{0}^{a} f(x)dx$,when $f(2a-x)=f(x)$ I got

$$\displaystyle\lim_{n \to \infty} \int_{o}^{\pi} \frac{\sin x}{1+\cos ^2(nx)} dx=2\displaystyle\lim_{n \to \infty} \int_{o}^{\pi/2} \frac{\sin x}{1+\cos ^2(nx)} dx$$ but I cannot proceed after that. Could someone provide me with some hint? Till now I have only done integration in terms of elementary functions. Any hint would be appreciated.

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  • $\begingroup$ GeoGebra seems to converge to $\approx 1.41$ for $n=400$ $\endgroup$ – Aniruddha Deb May 19 at 14:37
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    $\begingroup$ One approach is shown here. $\endgroup$ – Axion004 May 19 at 15:25
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Here are some big hints: \begin{align} \int_0^{\pi}\frac{\sin x}{1+\cos^2nx}\mathrm{d}x &= \frac{1}{n}\int_0^{n\pi}\frac{\sin(\theta/n)}{1+\cos^2\theta}\mathrm{d}\theta\\ &= \frac{1}{n}\sum_{k=0}^{n-1}\int_{k\pi}^{(k+1)\pi}\frac{\sin(\theta/n)}{1+\cos^2\theta}\mathrm{d}\theta\\ &=\frac{1}{n}\sum_{k=0}^{n-1}\int_{0}^{\pi}\frac{\sin\big(\frac{\psi+k\pi}{n}\big)}{1+\cos^2\psi}\,d\psi\\ &=\frac{1}{\pi}\int_0^{\pi}\frac{1}{1+\cos^2\psi}\Big[\frac{\pi}{n}\sum_{k=0}^{n-1}\sin\Big(\frac{\psi+k\pi}{n}\Big)\Big]\,d\psi \\ &\to\frac{1}{\pi}\int_0^{\pi}\frac{1}{1+\cos^2\psi}\,d \psi\cdot\int_0^{\pi}\sin t\,dt\\ &\ldots \end{align}

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  • $\begingroup$ I gave a link to the general result based on this idea in my answer. +1 $\endgroup$ – Paramanand Singh May 19 at 15:26
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    $\begingroup$ This is the same approach shown in the accepted answer to this question. $\endgroup$ – Axion004 May 19 at 15:27
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On the interval $[0,\pi]$ we have that $0\leq \cos^2(nx) \leq 1$, so rewrite the integral using a geometric series:

$$I_n = \int_0^\pi \frac{\sin x}{1+\cos^2(nx)}\:dx = \sum_{k=0}^\infty (-1)^k \int_0^\pi \sin x \cos^{2k}(nx)\:dx$$

Then use $\cos x = \frac{e^{ix}+e^{-ix}}{2}$ to turn the integral into a binomial series

$$I_n = \sum_{k=0}^\infty \left(-\frac{1}{4}\right)^k\sum_{l=0}^{2k} {2k \choose l} \int_0^\pi e^{i2nx(k-l)}\sin x\:dx$$

The integral can be further broken down into

$$\int_0^\pi e^{i2nx(k-l)}\sin x\:dx = \int_0^\pi \cos(2nx[k-l])\sin x\:dx + i\int_0^\pi \sin(2nx[k-l])\sin x\:dx$$

by Euler's formula. For all $k\neq l$ and large enough $n$, the functions are orthogonal on the interval $[0,\pi]$, so the integrals will be $0$, leaving the only surviving term as

$$I_n \to \sum_{k=0}^\infty \left(-\frac{1}{4}\right)^k \cdot {2k \choose k} \cdot 2 = \frac{2}{\sqrt{1+1}} = \sqrt{2}$$

from the Taylor series

$$\frac{1}{\sqrt{1-4x}} = \sum_{k=0}^\infty {2k \choose k} x^k$$

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  • $\begingroup$ Beautiful proof and a very direct one at that. +1 $\endgroup$ – Paramanand Singh May 19 at 15:24
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In this answer it is proved that if $f, g$ are Riemann integrable on $[0,T]$ and $g$ is periodic with period $T$ then $$\lim_{n\to\infty} \int_{0}^{T}f(x)g(nx)\,dx=\frac{1}{T}\left(\int_{0}^{T}f(x)\,dx\right)\left(\int_{0}^{T}g(x)\,dx\right)$$ For your problem we have $$f(x) =\sin x, g(x) =\frac{1}{1+\cos^2x}$$ and hence the desired limit is $$\frac{1}{\pi}\int_{0}^{\pi}\sin x\, dx\int_{0}^{\pi}\frac{dx}{1+\cos^2x}=\frac{4}{\pi}\int_{0}^{\pi}\frac{dx}{3+\cos x}=\frac{4}{\pi}\cdot\frac{\pi}{2\sqrt{2}}=\sqrt{2}$$

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