100 Followers Problem (Calculus)!

First of all, understand what the above theorem says. It's very simple to understand, for those who don't know it, and I expect a lot of people to know it.

Denote $P = \left(1+\frac{\sin\theta\sec^2\theta}{n}\right)^{\cos\theta} , Q=\left(1+\frac{\cos\theta}{n}\right)^{\cot\theta}, R=\left(1+\frac{\cot\theta}{n}\right)^{\sin\theta\sec^2\theta}$

Thus we have the integral as $\lim_{n \to \infty} \int_{\pi/4}^{\pi/3} n\ln{[PQR]}\ d\theta$

$= \displaystyle \lim_{n \to \infty} \int_{\pi/4}^{\pi/3} n\ln{(P)}\ d\theta +\displaystyle \lim_{n \to \infty} \int_{\pi/4}^{\pi/3} n\ln{(Q)}\ d\theta+ \displaystyle \lim_{n \to \infty} \int_{\pi/4}^{\pi/3} n\ln{(R)}\ d\theta$

$= \displaystyle \int_{\pi/4}^{\pi/3} \displaystyle \lim_{n \to \infty} n\ln{(P)}\ d\theta +\int_{\pi/4}^{\pi/3} \displaystyle \lim_{n \to \infty} n\ln{(Q)} \ d\theta+ \int_{\pi/4}^{\pi/3} \displaystyle \lim_{n \to \infty} n\ln{(R)}\ d\theta$

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Now:

$\displaystyle \int_{\pi/4}^{\pi/3} \displaystyle \lim_{n \to \infty} n\ln{(P)}\ d\theta = \displaystyle \int_{\pi/4}^{\pi/3} \displaystyle \lim_{n \to \infty} \frac{\ln{(1+\frac{\sec\theta \tan\theta}{n}})}{\frac{\sec\theta}{n}}\ d\theta$

$=\displaystyle \int_{\pi/4}^{\pi/3} \displaystyle \lim_{n \to \infty} \frac{\tan\theta\ln{(1+\frac{\sec\theta \tan\theta}{n}})}{\frac{\sec\theta\tan\theta}{n}}\ d\theta = \displaystyle \int_{\pi/4}^{\pi/3} \tan\theta \ d\theta$

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Similarly:

$\displaystyle \int_{\pi/4}^{\pi/3} \displaystyle \lim_{n \to \infty} n\ln{(Q)}\ d\theta$

$= \displaystyle \int_{\pi/4}^{\pi/3} \displaystyle \lim_{n \to \infty} \cot\theta\cos\theta\frac{\ln{(1+\frac{\cos\theta}{n}})}{\frac{\cos\theta}{n}}\ d\theta = \displaystyle \int_{\pi/4}^{\pi/3} \cot\theta\cos\theta \ d\theta$

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Similarly: $\displaystyle \int_{\pi/4}^{\pi/3} \displaystyle \lim_{n \to \infty} n\ln{(R)}\ d\theta$

$= \displaystyle \int_{\pi/4}^{\pi/3}\displaystyle \lim_{n \to \infty} \sec\theta\frac{\ln{(1+\frac{\cot\theta}{n}})}{\frac{\cot\theta}{n}}\ d\theta = \displaystyle \int_{\pi/4}^{\pi/3} \sec\theta \ d\theta$

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Therefore we obtain:

$\displaystyle \lim_{n \to \infty} \int_{\pi/4}^{\pi/3} n\ln{(P)}\ d\theta +\displaystyle \lim_{n \to \infty} \int_{\pi/4}^{\pi/3} n\ln{(Q)}\ d\theta+ \displaystyle \lim_{n \to \infty} \int_{\pi/4}^{\pi/3} n\ln{(R)}\ d\theta$

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$=\displaystyle \int_{\pi/4}^{\pi/3} \left(\tan\theta + \cot\theta\cos\theta + \sec\theta \right) \ d\theta$

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And solving this elementary definite integral gives us the value:

$\ln\left({\frac{(\sqrt{3} + 1)^2}{\sqrt{6}}}\right) +\left(\frac{1}{2} - \frac{1}{\sqrt{2}}\right)$

And therefore $\large{A+B+C+D+E = \boxed{15}}$ .