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?

Printable View

- Apr 22nd 2009, 08:38 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, 08:58 PMmollymcf2009
- Apr 22nd 2009, 09:06 PMflatwhitecrx
Wait, so how exactly will i subsitule that in?

Will I not have to do a limit? - Apr 22nd 2009, 09:14 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:

$\displaystyle \int^{\pi}_{0} tan(x) dx$

$\displaystyle = \int^{\pi}_{0} \frac{sinx}{cosx}$

Let $\displaystyle u = cosx$

$\displaystyle du = - sinx dx$

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

$\displaystyle u = cos(\pi) = -1$

$\displaystyle u = cos(0) = 1$

So your new integral is:

$\displaystyle -\int^{-1}_{1} \frac{1}{u} du$

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, 09:25 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, 09:37 PMChris L T521
It should be clear that the integral diverges.

when you go to integrate $\displaystyle \int_0^{\pi}\tan x\,dx$, you apply the substitution $\displaystyle u=\cos x$ to end up with the integral $\displaystyle \int_{-1}^1\frac1u\,du$. But this gives us $\displaystyle \left.\left[\ln u\right]\right|_{-1}^1$. Since $\displaystyle \ln u$ is only defined over $\displaystyle u>0$, $\displaystyle \left.\left[\ln u\right]\right|_{-1}^1\rightarrow\lim_{a\to0^+}\left.\left[\ln u\right]\right|_{a}^1=\lim_{a\to0^+}-\ln a=+\infty$

Thus, the integral diverges. - Apr 22nd 2009, 09:40 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:

$\displaystyle - \int^{\pi}_{0} \frac{1}{cos(x)}dx$

$\displaystyle = - \left[ln \left|cos(x)\right|\right]$

*Note the absolute value of your cos function!

So, $\displaystyle |cos(\pi)| = 1$ and $\displaystyle |cos(0)| = 1$

So, $\displaystyle ln(1) - ln(1) = 0 $ - Apr 22nd 2009, 09:48 PMChris L T521
Correct.

Quote:

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

$\displaystyle - \int^{\pi}_{0} \frac{1}{cos(x)}dx$

$\displaystyle = - \left[ln \left|cos(x)\right|\right]$

*Note the absolute value of your cos function!

Quote:

So, $\displaystyle |cos(\pi)| = 1$ and $\displaystyle |cos(0)| = 1$

So, $\displaystyle ln(1) - ln(1) = 0 $

**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, 10:13 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, 10:46 PMflatwhitecrx
Ok, I figured it out. Thank you guys.