How to find which expression is larger: $$e^{\pi\cdot \int_0^1\sqrt {\tan(x)}dx}$$ $$or$$ $$ \pi^{e\cdot \int_0^1\sin^2(x)dx}$$

This question seems bit complicated. What is the best way to start off analytically without using a calculator?

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    $\begingroup$ No integral except the integral $\int_{0}^{1}x^a\,dx$ that is trivial (it is $\frac{1}{a+1}$ for any $a>-1$.) $\endgroup$ – Jack D'Aurizio Apr 19 '17 at 9:48

Since on the interval $(0,1)$ we have $\sin(x)\leq x\leq\tan(x)$ and $2<e<3<\pi<4$,

$$ e^{\pi\int_{0}^{1}\sqrt{\tan x}\,dx} \geq e^{\pi \int_{0}^{1}\sqrt{x}\,dx} = e^{2\pi/3} \geq e^2 >4 $$ while $$ \pi^{e\int_{0}^{1}\sin(x)^2\,dx} \leq \pi^{e/3} \leq \pi < 4.$$

The comparison is not difficult, since the difference between such numbers is pretty large.

  • $\begingroup$ I think it is a very good solution! except knowing $e^2 \ge 7$ without calculator. However we can safely say that $e^2 \ge 4$ without using calculator. Hence the question. One expression evaluates to $\ge 4$ and the other $\le 4$ $\endgroup$ – user12345 Apr 19 '17 at 9:54
  • $\begingroup$ @user12345: $e\geq \left(1+\frac{1}{4}\right)^4$ works just as well. $\endgroup$ – Jack D'Aurizio Apr 19 '17 at 9:55
  • $\begingroup$ Thanks Jack! wonderful and short way to solve this problem $\endgroup$ – user12345 Apr 19 '17 at 9:56
  • $\begingroup$ @user12345: you're welcome. I have further simplified the answer, $2<e<3<\pi<4$ is enough. $\endgroup$ – Jack D'Aurizio Apr 19 '17 at 9:59
  • $\begingroup$ what a beautiful problem with short and cute solution. though looks scary at first look. $\endgroup$ – user12345 Apr 19 '17 at 13:45

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