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UDI MAGIC PROFESSIONAL CRACK VERSIONrar

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.Q:

$e^{ -pi iint frac{1}{x},dx} = frac{1}{x}+C$

I was working on a problem. I am aware that there are simpler methods, but it is the final solution to a problem I am using.
$$e^{ -pi iint frac{1}{x},dx}=
begin{cases}
phantom{ – }frac{1}{x}+C
&quad,text{ if }x
eq0,
\
-frac{pi}{2}
&quad,text{ if }x=0,
end{cases}$$
I found this solution below.

Here, he is utilizing the fact that $sin z=zcos z$ and $cos z=1-2sin^2frac{z}{2}$. I get the first part, but I don’t get the second part. Can somebody please explain why that is?

A:

The value of $int_{0}^{1}frac{1}{x},dx$ is $pi/2$. Therefore, $-frac{pi}{2}$ is the correct result.

A:

I suspect your professor had a typo. Perhaps he meant
$$e^{ -pi ileft(-pi^2right)}=color{red}{frac{pi}{x}+C}$$
which integrates to
$$-frac{pi^2}{2}$$
where $frac{pi^2}{2} https://www.raven-guard.info/pomodoro-app-3264bit-final-2022/

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