How do you find the integral of #int cscx dx# from pi/2 to pi?

This is an interesting intergal, we can find its antiderivative via considering a valid substitution.

To find the integral of $\int_{\frac{\pi}{2}}^{\pi} \csc(x) \, dx$, first, recall that $\csc(x) = \frac{1}{\sin(x)}$. Using this substitution, the integral becomes:

$\int_{\frac{\pi}{2}}^{\pi} \frac{1}{\sin(x)} \, dx$

Now, make the substitution $u = \cos(x)$ and $du = -\sin(x) \, dx$, which gives:

$\int_{0}^{-1} -\frac{1}{u} \, du$

After integrating, you'll get:

$-\ln|u| \bigg|_{0}^{-1}$

Finally, evaluate at the limits:

$-\ln|-1| - (-\ln|0|)$

Since $\ln|0|$ is undefined, the integral is divergent.

To find the integral of $\int_{\frac{\pi}{2}}^{\pi} \csc(x) \, dx$, we first rewrite $\csc(x)$ as $\frac{1}{\sin(x)}$. Then, we integrate $\frac{1}{\sin(x)}$ with respect to $x$ from $\frac{\pi}{2}$ to $\pi$. This integral is equal to $\ln|\csc(x) + \cot(x)|$, evaluated from $\frac{\pi}{2}$ to $\pi$. Substituting $\pi$ and $\frac{\pi}{2}$ into the antiderivative and subtracting the result at $\frac{\pi}{2}$ from the result at $\pi$ gives the final answer. The integral evaluates to $\ln|(-1) + 0| - \ln|\infty + 1|$, which simplifies to $-\ln(1) - \ln(\infty)$. Since $\ln(1) = 0$ and $\ln(\infty) = \infty$, the final result is $-\infty$.