Will you keep the value?

Taking limits, $x$ is a constant in first expression, and $t$ is the variable. As $t \rightarrow 0$ , the term $\frac{1}{t^{2}}\rightarrow \pm \infty$ ,depending on $sgn(x)$ , except for $x=0$ , on which its undefined. Then, we have:

• $\displaystyle x>0, \ L=lim_{t \rightarrow 0} \ arccot\left(\frac{x}{t^{2}}\right) \implies L=- \pi;$

• $\displaystyle x<0, \ L=lim_{t \rightarrow 0} \ arccot\left(\frac{x}{t^{2}}\right) \implies L= 0;$

• $\displaystyle x=0, \ L=lim_{t \rightarrow 0} \ arccot\left(\frac{x}{t^{2}}\right) \implies \text{L is undefined.}$

Heading back to $f(x)$ , it becomes:

• $\displaystyle x>0 \implies f(x)=sin(-2x)=-sin(2x);$

• $\displaystyle x<0 \implies f(x)=0;$

• $\displaystyle x=0 \implies \text{f(x) is undefined.}$

There's a theorem in calculus which states that for a finite number of discontinuities, the area under the function with these discontinuities remains unchanged, as if there really were the remaining points.

Since the function is null before zero, it's only needed to integrate the function $f(x)=-sin(2x)$ at $(0, \frac{\pi}{2})$ . Integrating gives:

$\displaystyle \int_{0}^{\frac{\pi}{2}} -sin(2x) \ dx=\left[ \frac{cos(2x)}{2} \right] \bigg \rvert_{0}^{\frac{\pi}{2}}=\boxed{-1}$