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

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.

• GeoGebra seems to converge to $\approx 1.41$ for $n=400$ – Aniruddha Deb May 19 at 14:37
• One approach is shown here. – Axion004 May 19 at 15:25

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}

• I gave a link to the general result based on this idea in my answer. +1 – Paramanand Singh May 19 at 15:26
• This is the same approach shown in the accepted answer to this question. – Axion004 May 19 at 15:27

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

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

• Beautiful proof and a very direct one at that. +1 – Paramanand Singh May 19 at 15:24

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 $$\frac{1}{\pi}\int_{0}^{\pi}\sin x\, dx\int_{0}^{\pi}\frac{dx}{1+\cos^2x}=\frac{4}{\pi}\int_{0}^{\pi}\frac{dx}{3+\cos x}=\frac{4}{\pi}\cdot\frac{\pi}{2\sqrt{2}}=\sqrt{2}$$