# How do you find the integral of #(5 csc x dx)# from #[pi/2,pi]#?

The integral deviates.

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To find the integral of ( \int 5 \csc(x) , dx ) from ( \frac{\pi}{2} ) to ( \pi ):

First, recall the integral of ( \csc(x) ):

[ \int \csc(x) , dx = -\ln|\csc(x) + \cot(x)| + C ]

Now, integrate ( 5 \csc(x) ):

[ \int 5 \csc(x) , dx = -5\ln|\csc(x) + \cot(x)| + C ]

To find the definite integral from ( \frac{\pi}{2} ) to ( \pi ), evaluate the antiderivative at the upper and lower bounds and subtract:

[ \int_{\frac{\pi}{2}}^{\pi} 5 \csc(x) , dx = \left[ -5\ln|\csc(x) + \cot(x)| \right]_{\frac{\pi}{2}}^{\pi} ]

Now, substitute the upper and lower bounds:

[ \left[ -5\ln|\csc(\pi) + \cot(\pi)| \right] - \left[ -5\ln|\csc\left(\frac{\pi}{2}\right) + \cot\left(\frac{\pi}{2}\right)| \right] ]

[ = \left[ -5\ln|(-1) + 0| \right] - \left[ -5\ln|\infty + 1| \right] ]

[ = \left[ -5\ln(1) \right] - \left[ -5\ln(\infty) \right] ]

Since ( \ln(\infty) ) approaches infinity:

[ = -5(0) + 5(\infty) ]

The integral is divergent because the function ( 5 \csc(x) ) has vertical asymptotes at the limits of integration, and the logarithmic function approaches infinity near these asymptotes.

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To find the integral of (5 \csc(x) , dx) from (x = \frac{\pi}{2}) to (x = \pi), you can follow these steps:

- Rewrite (5 \csc(x)) as (\frac{5}{\sin(x)}).
- Recognize that (\frac{1}{\sin(x)}) is the same as (\csc(x)).
- Since the integral of (\csc(x)) is (-\ln|\csc(x) + \cot(x)| + C), integrate (\frac{5}{\sin(x)}) with respect to (x) using the same method.

The definite integral will be evaluated from (x = \frac{\pi}{2}) to (x = \pi).

So, the integral of (5 \csc(x) , dx) from (x = \frac{\pi}{2}) to (x = \pi) is:

[-5\ln\left|\csc(x) + \cot(x)\right| \Bigg|_{\frac{\pi}{2}}^{\pi}]

[= -5\ln\left|\csc(\pi) + \cot(\pi)\right| + 5\ln\left|\csc\left(\frac{\pi}{2}\right) + \cot\left(\frac{\pi}{2}\right)\right|]

[= -5\ln\left|\frac{1}{\sin(\pi)} + \cot(\pi)\right| + 5\ln\left|\frac{1}{\sin(\frac{\pi}{2})} + \cot(\frac{\pi}{2})\right|]

[= -5\ln\left|\frac{1}{0} + 0\right| + 5\ln\left|\frac{1}{1} + 0\right|]

Since (\sin(\pi) = 0) and (\cot(\pi) = 0), the first term results in an indeterminate form. We cannot directly evaluate this term.

However, (\sin\left(\frac{\pi}{2}\right) = 1) and (\cot\left(\frac{\pi}{2}\right) = 0), so the second term simplifies to:

[= -5\ln\left|\frac{1}{1} + 0\right| + 5\ln\left|\frac{1}{1} + 0\right|]

[= -5\ln(1) + 5\ln(1)]

[= 0]

Hence, the integral of (5 \csc(x) , dx) from (x = \frac{\pi}{2}) to (x = \pi) evaluates to (0).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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