find the minimum value of "a" such that the following integral holds good, [ ] represents the greatest integer function

Question

Finding minimum value of $a$ such that following integrals holds

$$\int^{a}_{0}\bigg\lfloor \tan^{-1}(\sqrt{x})\bigg\rfloor dx=\int^{a}_{0}\bigg\lfloor \cot^{-1}(\sqrt{x})\bigg\rfloor dx$$

Answer 1

Notice that this is equivalent to:

$$\int_{\tan^2(1)}^a [\arctan(\sqrt{x}))] dx=\int_{0}^{a} [\text{arccot}(\sqrt{x}))] dx$$

Since:

$$a \geq \tan^2(1) \geq \cot^2(1)$$

So the question becomes: $$\int_{\tan^2(1)}^a [\arctan(\sqrt{x})] dx=\int_{0}^{\cot^2(1)} [\text{arccot}(\sqrt{x})] dx$$

Since these areas are basically rectangles we'll have:

$$(a-\tan^2(1))=\cot^2(1) \Rightarrow a=\tan^2(1)+\cot^2(1)\approx2.837...$$

:)