Value of $a$ for which two integrals are equal Determine $a$ such that $$\int _0 ^a ([\arctan \sqrt {x}]dx=\int _0 ^a  [\frac {\pi}{2}-\arctan\sqrt {x}]dx$$ where $[.] $ is greatest  integer function so i first took the first integral to get integral as $a [\arctan (a)] $ as I think derivative of integer function is 0 as its always an integer. But I dont know how to proceed. 
 A: $\tan^{-1} \sqrt{x}$ is defined for $x\ge 0$, so we need to consider $a\ge 0$ only.



*

*$\tan^{-1} \sqrt{x}$ is an increasing function and


$$0\le \tan^{-1} \sqrt{x} \le \frac{\pi}{2} < 2$$
\begin{align*}
  \left \lfloor \tan^{-1} \sqrt{x} \right \rfloor &=
   \left \{
    \begin{array}{lr}
      0  \, , & 0\le x < \tan^2 1 \\
      1  \, , & x \ge \tan^2 1
  \end{array}
  \right. \\[5pt]
  \int_0^a
  \left \lfloor
    \tan^{-1} \sqrt{x}
  \right \rfloor \, dx &=
  \left \{
    \begin{array}{lr}
      0  \, , & 0\le a < \tan^2 1 \\
      a-\tan^2 1  \, , & a \ge \tan^2 1
    \end{array}
  \right.
\end{align*}



*

*$\dfrac{\pi}{2}-\tan^{-1} \sqrt{x}$ is a decreasing function and


$$0\le \frac{\pi}{2}-\tan^{-1} \sqrt{x} \le \frac{\pi}{2} < 2$$
\begin{align*}
  \left \lfloor
    \frac{\pi}{2}-\tan^{-1} \sqrt{x}
  \right \rfloor &=
  \left \{
    \begin{array}{lr}
      1  \, , & 0\le x < \cot^2 1 \\
      0  \, , & x \ge \cot^2 1
    \end{array}
  \right. \\[5pt]
  \int_0^a
  \left \lfloor
    \frac{\pi}{2}-\tan^{-1} \sqrt{x}
  \right \rfloor \, dx &=
  \left \{
    \begin{array}{lr}
      a  \, , & 0\le a < \cot^2 1 \\
      \cot^2 1  \, , & a \ge \cot^2 1
    \end{array}
  \right.
\end{align*}

Note that $\cot^2 1 <\tan^2 1$,
$$\int_0^a
  \left \lfloor
    \tan^{-1} \sqrt{x}
  \right \rfloor \, dx =
\int_0^a
  \left \lfloor
    \frac{\pi}{2}-\tan^{-1} \sqrt{x}
  \right \rfloor \, dx$$

$$\implies a=0 \quad \text{or} \quad a=\tan^2 1+\cot^2 1$$

