How to show the inequality for $\int_{0}^{\pi/4}\tan^n(x) dx$

Question

I can't seem to figure out the inequality option, I substituted n=2 for the answer bit is there any elegant way of getting it?

Answer 1

Since $\tan x <1 ~ \forall x \in \left(0,\dfrac{\pi}{4}\right)$, We can say that -

$$\int_{0}^{\pi/4}\tan^nx <\int_{0}^{\pi/4}\tan^{n-2}x$$

And also that

$$\int_{0}^{\pi/4}\tan^nx > \int_{0}^{\pi/4}\tan^{n+2}x$$

Hence

$$I_n+I_{n-2}> I_n+I_n=2I_n \implies \color{blue}{\frac{1}{n-1}>2I_n} \tag 1$$

and

$$I_n+I_{n+2}< I_n+I_n=2I_n \implies\color{blue}{\frac{1}{n+1}<2I_n} \tag2$$

Combinig $(1)$ and $(2)$, we get

$$\boxed{\color{green}{\frac{1}{n+1}<2I_n <\frac{1}{n-1}}}$$