I Was Vey Amazed At The Solution 4

When integrated with respect to x, $\displaystyle\implies f(x) = ( \tan x - x) + \left(\int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} f(x) \ dx \right)x + C$

Now,

$\displaystyle \begin{aligned} f\left(\dfrac{\pi}{4}\right) & = \left(\tan \left(\dfrac{\pi}{4}\right) - \dfrac{\pi}{4}\right) + \left[ \int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} f(x) \ dx \right]\cdot \dfrac{\pi}{4} + C = -\dfrac{\pi}{4} \\ \implies C & = -1 - \left[ \int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} f(x) \ dx \right]\cdot \frac{\pi}{4} \end{aligned}$

Hence, $\displaystyle f(x) = ( \tan x - x) + \left[\int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} f(x) \ dx\right]x - \left(1 + \left( \dfrac{\pi}{4} \cdot \int \limits_{\frac{-\pi}{4}}^{\frac{\pi}{4}} f(x) \ dx \right) \right)$

integrating both sides with respect to $x$ and whose limits is $\cfrac{-\pi}{4}$ to $\cfrac{\pi}{4}$ and let $\displaystyle A = \int \limits_{\frac{-\pi}{4}}^{\frac{\pi}{4}} f(x) \ dx$

$\displaystyle \implies A = \left. \left( -\ln |\cos x| - \dfrac{x^2}{2} \right) \right|_{\frac{-\pi}{4}}^{\frac{\pi}{4}} + \left. \left(\dfrac{Ax^2}{2}\right) \right|_{\frac{-\pi}{4}}^{\frac{\pi}{4}} - \left. \left( 1 + A\dfrac{\pi}{4} \right)x \right|_{\frac{-\pi}{4}}^{\frac{\pi}{4}}$

As both $\displaystyle -\ln |\cos x| - \dfrac{x^2}{2}$ and $\displaystyle \dfrac{Ax^2}{2}$ are even functions, hence, when they are evaluated at limits $-a$ and $a$ , their result will be 0.

$\displaystyle \implies A = - \left. \left( 1 + A\dfrac{\pi}{4} \right)x \right|_{\frac{-\pi}{4}}^{\frac{\pi}{4}}$

$\vdots$

$\displaystyle A = - \left( 1 + A\dfrac{\pi}{4} \right) \left(\dfrac{\pi}{4} - \dfrac{-\pi}{4} \right) \\ A = - \left( 1 + A\dfrac{\pi}{4} \right) \left( \dfrac{\pi}{2} \right)$

$\vdots$ $\displaystyle A = - \dfrac{\pi}{2} - A\dfrac{\pi^2}{8} \\ A\left(1 + \dfrac{\pi^2}{8} \right) = - \dfrac{\pi}{2} \\ A = - \dfrac{ \dfrac{\pi}{2}}{\left(1 + \dfrac{\pi^2}{8} \right)} = - \dfrac{4\pi}{8+\pi^2}$

$\vdots$

$\begin{aligned} \displaystyle \implies \left \lceil -1000 \int \limits_{\dfrac{-\pi}{4}}^{\dfrac{\pi}{4}} f(x) \ dx \right \rceil = \left \lceil -1000A \right \rceil \\ = \left \lceil -1000 \left(-\dfrac{4\pi}{8+\pi^2}\right) \right \rceil = \boxed{704} \end{aligned}$

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Answer to the Bonus Question:

Since $\displaystyle f(x) = ( \tan x - x) + \left[\int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} f(x) \ dx\right]x - \left(1 + \left( \dfrac{\pi}{4} \cdot \int \limits_{\frac{-\pi}{4}}^{\frac{\pi}{4}} f(x) \ dx \right) \right)$ and

$\displaystyle A = \int \limits_{\frac{-\pi}{4}}^{\frac{\pi}{4}} f(x) \ dx = - \dfrac{4\pi}{8+\pi^2}$

$\displaystyle \implies f(x) = ( \tan x - x) - \left( \dfrac{4\pi}{8+\pi^2} \right)x - \left( 1 - \left( \dfrac{\pi}{4} \cdot \dfrac{4\pi}{8+\pi^2} \right) \right)$

or simply, $\displaystyle \boxed{f(x) = ( \tan x - x) - \left( \dfrac{4\pi}{8+\pi^2} \right)x - \left( 1 - \dfrac{\pi^2}{8+\pi^2} \right)}$