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So recently I found this integral: $$\lim_{n \rightarrow \infty} \int_{0}^{\pi/3} \frac{\sin^{n}x}{\sin^{n}x+\cos^{n}x}dx$$

I know the answer should be $ \frac{\pi}{12} $ and I saw it can be solved using the Dominated Convergence Theorem. I managed to get the integral to this form: $$\lim_{n \rightarrow \infty} \frac{\pi}{3}- \int_{0}^{\pi/3} \frac{1}{1+\operatorname{tg}^{n}x}dx$$ The new integral should be $ \frac{\pi}{4} $. But I can't find the function which bounds the function inside the integral. Can you explain me how can I find the answer using DCT? Thanks in advance.

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Hint. One may just observe that, as $n \to \infty$, $$ \begin{align} \int_{0}^{\pi/3} \frac{1}{1+\tan^{n}x}dx&=\int_{0}^{\pi/4} \underbrace{\frac{1}{1+\tan^{n}x}}_{\to \,\color{red}{1}\, \text{using} \,|\tan x|<1}dx+\int_{\pi/4}^{\pi/3} \underbrace{\frac{1}{1+\tan^{n}x}}_{\to \,\color{red}{0}\, \text{using} \,|\tan x|>1}dx \\& \to \frac \pi4+\color{red}{0}.\end{align} $$ Then apply the DCT appropriately.

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Since for all n from here we have Integrate $\int_0^{\pi/2} \frac{1}{1+\tan^\alpha{x}}\,\mathrm{d}x$ $$\int_0^{\pi/2} \frac{1}{1+\tan^n x}\,\mathrm{d}x =\frac{π}{ 4}$$

Enforcing $x= \pi/2-t$. we get, $$I= \int_0^{\pi/3} \frac{1}{1+\tan^n x}\,dx =\int_ {\pi/6} ^{\pi/2} \frac{1}{1+\tan^n(\pi/2-t)}\,dt \\=\int_ {\pi/6} ^{\pi/2} \frac{\tan^n(t)}{1+\tan^n(t)}\,dt = \frac{\pi}{3}-\int_ {\pi/6} ^{\pi/2} \frac{1}{1+\tan^n(t)}\,dt\\= \frac{\pi}{3}-\int_ {0} ^{\pi/2} \frac{1}{1+\tan^n(t)}\,dt+\int_ {0} ^{\pi/6} \frac{1}{1+\tan^n(t)}\,dt \\= \frac{\pi}{3}-\frac{\pi}{4}+\int_ {0} ^{\pi/6} \frac{1}{1+\tan^n(t)}\,dt\to \color{red}{\frac{\pi}{3}-\frac{\pi}{4}+\frac{\pi}{6} =\frac{\pi}{4}}$$

Since $\tan $ is continuous on the compact set $[0,π/6]$ and for $\in [0,π/6]$we have

$$ 0\le \tan^n x\le \tan^n(π/6)=\frac{1}{(\sqrt{3})^n}\to 0$$

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Take the constant function $1$. If $x\in\left[0,\frac\pi3\right]$, then$$1+\tan^nx\geqslant 1$$and therefore$$\frac1{1+\tan^nx}\leqslant1.$$

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