How do I evaluate #int_(pi/2)^pi5 csc(x) dx#?

Question

Christopher Allers · Accepted Answer

I am not sure you can solve this...I mean, your function #5csc(x)=5/sin(x)#, when #x=pi# goes towards infinity...so, the área underneath it cannot be finite!

I gave a shot to the integral anyway:

If you now substitute the value of #pi# for your integral you hit the problem.

Evelyn Agard · Answer

undefined To evaluate the integral $\int_{\frac{\pi}{2}}^{\pi} 5 \csc(x) \, dx$, you can use the following steps:

1. Rewrite $\csc(x)$ as $\frac{1}{\sin(x)}$.
2. Then, the integral becomes $\int_{\frac{\pi}{2}}^{\pi} 5 \cdot \frac{1}{\sin(x)} \, dx$.
3. Use the property $\int \frac{1}{\sin(x)} \, dx = -\ln|\csc(x) + \cot(x)| + C$.
4. Apply the bounds of integration to find the definite integral.

So, the integral evaluates to $5(-\ln|\csc(\pi) + \cot(\pi)| + \ln|\csc(\frac{\pi}{2}) + \cot(\frac{\pi}{2})|)$.

Note that $\csc(\pi) = \frac{1}{\sin(\pi)} = \frac{1}{0}$, which is undefined. Similarly, $\cot(\pi) = \frac{\cos(\pi)}{\sin(\pi)} = \frac{-1}{0}$, also undefined. However, $\csc(\frac{\pi}{2}) = \frac{1}{\sin(\frac{\pi}{2})} = \frac{1}{1} = 1$ and $\cot(\frac{\pi}{2}) = \frac{\cos(\frac{\pi}{2})}{\sin(\frac{\pi}{2})} = \frac{0}{1} = 0$.

So, the integral evaluates to $5(-\ln|0 + 0| + \ln|1 + 0|)$, which simplifies to $5(-\ln|0| + \ln|1|)$.
The natural logarithm of 0 is undefined, so this integral is also undefined.