I want to evaluate the following integral
$$\int^1_0 \frac{\sin(\tan(x))}{x} \ dx $$
I found here that in the improper case it's possible to use the Lobachevsky Integral Formula. Is there anything similar in this case?
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1$1$ does not seem to be a 'sensible' bound for a function involving $\tan$, so basically you are trying to get the antiderivative of $\frac{\sin(\tan(x))}{x}$. In the (great) answer that you linked, there was a lot of use of the fact that the integral was from $0$ to $\infty$ (in order to use the expansion for $\frac{1}{\tan(x)}$ and to use the Feynman trick on $\frac{\sin(at)}{t(1+t^2)}$, which doesn't work anymore if you are not integrating on $\mathbb{R}_+$). Do you have reason to think that a closed form can be obtained for your integral from $0$ to $1$ (or really from $0$ to any $t$)? – charmd Jun 03 '23 at 07:03
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@charmd thanks for your answer. Well no, I have no reason to think that – user967210 Jun 03 '23 at 07:09