Definate integral calculation difficulty

**Furyan** · January 25th 2013, 03:05 PM

Hello,

The question is to evaluate the following integral.

$\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}\ 7 \cot^2(7x) dx$

by writing it as

$7\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}\ \csc^2(7x) - 1 dx$

I got

$-\cot(7x) -7x$

which I think is right, my difficulty seems to be with evaluating. The answer is given as $1+\dfrac{7\pi}{4}$ , but I keep getting $-1 - \dfrac{7\pi}{4}$ .

I'd be very grateful if someone would check this.

Thank you.

**ILikeSerena** · January 25th 2013, 03:19 PM

Hi Furyan!

You have a couple of singularities in there.
Perhaps you could split the integral in such a way you can take care of these singularities?

**Furyan** · January 25th 2013, 03:31 PM

Originally Posted by ILikeSerena

Hi Furyan!

You have a couple of singularities in there.
Perhaps you could split the integral in such a way you can take care of these singularities?

Hi ILikeSerena, I certainly feel very close to a black hole, but I'm afraid I can't resist the field strength

. I figure since I've got the integral it should just be plain sailing. What do you mean by singularities?

Thank you

**ILikeSerena** · January 26th 2013, 01:22 AM

Originally Posted by Furyan

Hi ILikeSerena, I certainly feel very close to a black hole, but I'm afraid I can't resist the field strength

. I figure since I've got the integral it should just be plain sailing. What do you mean by singularities?

Thank you

Actually, now that I'm taking a better look, it's even worse. Your integral does not exist.
Your given answer is wrong.

The problem is that $\cot 7x$ is not defined for all the points in the range.
It has so called singularities in $x=0, \frac \pi 7, \frac {2\pi} 7, \frac {3\pi} 7$ .

It means that you have to split your integral up at these singularities into so called improper integrals (see included wiki link).
Your integral expression $7 \cot^2(7x)$ has infinitely high mountains in these singularities.
Now it could still be possible that these mountains have a finite area, but in this case they don't.

If we split up your integral at the singularities, we get:

$\int_{\pi / 4}^{\pi / 2} 7 \cot^2(7x) dx = \int_{\pi / 4}^{2\pi / 7} 7 \cot^2(7x) dx + \int_{2\pi / 7}^{3\pi / 7} 7 \cot^2(7x) dx + \int_{3\pi / 7}^{\pi / 2} 7 \cot^2(7x) dx$

Let's pick the first term:

$\int_{\pi / 4}^{2\pi / 7} 7 \cot^2(7x) dx$

$= \lim_{b~ \uparrow~ {2\pi / 7}} \int_{\pi / 4}^{b} 7 \cot^2(7x) dx$

$= \lim_{b~ \uparrow~ {2\pi / 7}} (-\cot(7x)-7x)|_{\pi / 4}^b$

$=\text{does not exist}$

This improper integral does not exist, because $\cot(7\cdot {2\pi \over 7}) = \cot(2 \pi)$ is not defined.

None of these improper integrals exist, so your integral does not exist either $\Box$ .

**Furyan** · January 26th 2013, 10:54 AM

Originally Posted by ILikeSerena

Actually, now that I'm taking a better look, it's even worse. Your integral does not exist.
Your given answer is wrong.

Hi IlikeSerena,

Thank you for taking the time to post such a comprehensive reply. I now understand that the integral does not exist because you cannot integrate across a discontinuity. I also now see that it's important to sketch a graph when finding a definite integral and doing so helped me to see why I was getting a couple of other questions wrong. One thing that was confusing me was that when evaluating:

$\dfrac{1}{\tan\tfrac{7\pi}{2}}$ I got an error

but when evaluating

$\dfrac{\cos\tfrac{7\pi}{2}}{\sin\tfrac{7\pi}{2}}$ I got an answer of 0

I now have to evaluate:

$\int_0^2\dfrac{5 - 2x}{(x - 1)(2x + 1)}$

Integrating I get $\ln(x - 1) - 2\ln(2x + 1)$

The given answer is $\ln\dfrac{1}{25}$

Am I right in thinking this integral does not exist either since $y = \dfrac{1}{x - 1}$ is not defined when $x = 1?$

Thank you

**ILikeSerena** · January 26th 2013, 11:27 AM

Originally Posted by Furyan

Hi IlikeSerena,

Thank you for taking the time to post such a comprehensive reply. I now understand that the integral does not exist because you cannot integrate across a discontinuity. I also now see that it's important to sketch a graph when finding a definite integral and doing so helped me to see why I was getting a couple of other questions wrong.

Good!

One thing that was confusing me was that when evaluating:

$\dfrac{1}{\tan\tfrac{7\pi}{2}}$ I got an error

but when evaluating

$\dfrac{\cos\tfrac{7\pi}{2}}{\sin\tfrac{7\pi}{2}}$ I got an answer of 0

The point is that you cannot divide by zero (or something bad happens to the universe, like black holes and singularities and stuff

).

The tan function is actually a division of the sine by the cosine.
It does not exist if the cosine is zero, which is the case at 7pi/2, or more generally at pi/2 + k pi, where k is an integer.

When the tan does not exist, you can't take its inverse any more, even if that would have existed as in your case.

So $1 \over \displaystyle ({\sin 7\pi/2 \over \cos 7\pi/2})$ does not exist, because it contains a division by zero.
But ${\cos 7\pi/2 \over \sin 7\pi/2}$ does exist, since there is no division by zero.

I now have to evaluate:

$\int_0^2\dfrac{5 - 2x}{(x - 1)(2x + 1)}$

Integrating I get $\ln(x - 1) - 2\ln(2x + 1)$

The given answer is $\ln\dfrac{1}{25}$

Am I right in thinking this integral does not exist either since $y = \dfrac{1}{x - 1}$ is not defined when $x = 1?$

Thank you

Yes, you are correct. The integral does not converge due to the singularity at x=1.

Btw, your anti-derivative is not entirely correct, since it should also be defined for x < 1.
You lost that somewhere in your integration.
The anti-derivative you have only works for the part to the right of the singularity.
You would probably need to find another anti-derivative for the left part.

**Furyan** · January 26th 2013, 12:38 PM

Yes, you are correct. The integral does not converge due to the singularity at x=1.

Btw, your anti-derivative is not entirely correct, since it should also be defined for x < 1.
You lost that somewhere in your integration.
The anti-derivative you have only works for the part to the right of the singularity.
You would probably need to find another anti-derivative for the left part.

Thank you very much. Yes I can see that the anti-derivative only works for the part to the right of the singularity. For now I have no idea how I would go about finding another one for the left part, but I will think about it.

Thanks again for all your help.

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