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My question here is related to this question , but my question is to know if the below identity is true :

$$\lim_{n\to \infty}\int_0^{\pi/2} \frac{x^{\frac 1n}+\sin x}{\tan x+x^{\frac 1n}} dx=\int_0^{\pi/2} \frac{1+\sin x}{\tan x+1} dx=1.62...$$ .for The RHS of this identity the value of this integral assumed by wolfram alpha to be $1.62..$ with a closed form.

Note: The LHS of this identity it's seems converges for every positive integer $n\geq 1$ over the range $(0,\frac \pi2)$.

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    $\begingroup$ Are you familiar with the Dominated Convergence Theorem? $\endgroup$ – Mark Viola Apr 13 '18 at 14:29
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Noting $$ 0\le \sin x\le x, \tan x\ge x, x\in[0,\frac\pi2] $$ one has $$ 0\le\frac{x^{\frac1n}+\sin x}{\tan x+x^{\frac1n}}\le\frac{x^{\frac1n}+ x}{x+x^{\frac1n}}=1$$ which implies $$ \lim_{n\to\infty} \int_0^{\pi/2}\frac{x^{\frac1n}+\sin x}{\tan x+x^{\frac1n}}dx=\int_0^{\pi/2}\lim_{n\to\infty}\frac{x^{\frac1n}+\sin x}{\tan x+x^{\frac1n}}dx=\int_0^{\pi/2}\frac{1+\sin x}{\tan x+1}dx $$ by using the DCT. Now under $\tan(\frac x2)\to x$, one has $$ \int_0^{\pi/2}\frac{1+\sin x}{\tan x+1}dx=2\int_0^1\frac{(2x+1)(x^2-1)}{(x^2+1)(x^2-2x-1)}dx$$ which can be handled by partial fractions easily.

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