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How to prove $$\frac{1}{2n+2}<\int_0^{\frac{\pi}{4}}\tan^nx\,dx< \frac{1}{2n}$$ Set $A_n=\int_0^{\frac{\pi}{4}}\tan^nx\,dx$, then we have $A_n+A_{n+2}=\frac{1}{n+1}$ and we have $A_{n+2} < A_n$ ,so we can get $$\frac{1}{2n+2}< \int_0^{\frac{\pi}{4}}\tan^nx\,dx < \frac{1}{2n-2}$$ But how to show that$$\int_0^{\frac{\pi}{4}}\tan^nx\,dx < \frac{1}{2n}$$

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Change variable to $t = \tan x$, we have

$$I_n \stackrel{def}{=}\int_0^{\pi/4} \tan^n x dx = \int_0^1 \frac{t^n}{1+t^2} dt$$

Notice for $t \in (0,1)$, we have $\frac{1 + t^2}{2} < 1$. This implies

$$I_n > \int_0^1 \frac{t^n}{1+t^2}\cdot\frac{1+t^2}{2} dt = \frac12\int_0^1 t^n dt = \frac{1}{2(n+1)}$$

On the other direction, AM $\ge GM$ tell us $t = \sqrt{1 \cdot t^2} \le \frac{1+t^2}{2}$ and the inequality is strict when $t \ne 1$. This leads to

$$I_n = \int_0^1 \frac{t^{n-1}}{1+t^2} t dt < \int_0^1 \frac{t^{n-1}}{1+t^2}\cdot \frac{1+t^2}{2} dt = \frac12 \int_0^1 t^{n-1} dt = \frac{1}{2n}$$

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For $n=0$ there is nothing to prove. For $n=1$, this is $\log\sqrt{2}<\frac12$, which is obviously true.

We have $$ \tan^{n+1}x<\frac12(\tan^n x+\tan^{n+2}x)\text{ for }x\in(0,\pi/4) \tag{1} $$ because $y\in\mathbb{R}^+\mapsto c^y\in\mathbb{R}^+$ is strictly convex for $c>0$. So you get $A_{n+1}<\frac1{2(n+1)}$, or equivalently $A_n<\frac1{2n}$.

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