I have the integral of tanx dx integrating from 0 to pi. Because of the asymptote at pi/2 I split the integral up but when I work it all out i get infinity - infinity? which is undefined. How could I rewrite this to make it work?

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- Apr 22nd 2009, 09:38 PMflatwhitecrxLimit question
I have the integral of tanx dx integrating from 0 to pi. Because of the asymptote at pi/2 I split the integral up but when I work it all out i get infinity - infinity? which is undefined. How could I rewrite this to make it work?

- Apr 22nd 2009, 09:58 PMmollymcf2009
- Apr 22nd 2009, 10:06 PMflatwhitecrx
Wait, so how exactly will i subsitule that in?

Will I not have to do a limit? - Apr 22nd 2009, 10:14 PMmollymcf2009
Oops! Sorry I just noticed that you had limits in there, sorry. The easiest thing to do is to change your limits of integration so that you can lust evaluate them at u instead of having to replace u with the original substitution. Here is what you do:

So we have:

Let

Then change your limits by plugging in the old limits into your equation for u:

So your new integral is:

Integrate and just evaluate at those new limits. Just integrate 1/u the same way you would integrate 1/x

Does that help? - Apr 22nd 2009, 10:25 PMflatwhitecrx
well, considering I can't remember ever doing that in class, I'm not sure he would accept me doing that. It says evaluate the integral or show that it diverges. I just confused because I can't think of any way to rewrite it so that the it doesn't come out to be undefined. You know?

- Apr 22nd 2009, 10:37 PMChris L T521
- Apr 22nd 2009, 10:40 PMmollymcf2009
Just so you know, if you haven't learned u-substitution yet, you will very soon and you will use it all the time. There is no way around it...

The integral of tan(x) = -ln|cos(x)|which is defined for all x except where tan(x) itself is undefined.

If you plug in cos(x) back into your equation for u and then integrate, you get:

*Note the absolute value of your cos function!

So, and

So, - Apr 22nd 2009, 10:48 PMChris L T521
Correct.

Quote:

If you plug in cos(x) back into your equation for u and then integrate, you get:

*Note the absolute value of your cos function!

Quote:

So, and

So,

**area**is infinite, but its**net****area**is zero... ><

And then again, it comes down to whether or not we're looking for**net area**or**area**... XD - Apr 22nd 2009, 11:13 PMCalculus26
You can't integrate over a singularity

you have to integrate this from 0 to pi/2 and then pi/2 to pi

If either integral diverges the integral over the entire interval diverges.

You know inf -inf is indeterminate not necessarily 0

for eg lim (x-ln(x)) is inf

So there is no "net area" of 0-- the integral diverges period - Apr 22nd 2009, 11:46 PMflatwhitecrx
Ok, I figured it out. Thank you guys.