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?
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?
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,
Correct.
. !!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!
This is "valid", but when you plot the tangent fuction on the interval , you see that its area is infinite, but its net area is zero... ><So, and
So,
And then again, it comes down to whether or not we're looking for net area or area... XD
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