Integration of functions: problem with borders

[Gamma]

I have couple of problems that I am stuck with.

Hi,

Integration (tan(2x) dx limit pi/4 to 3 pi/4

When I intergrate this, I get ln [1/(sqrt(cos2x))]

So when put the limits, we have zero in the denominator. What do I do?
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The other problem is,

integration (1/(t-3) dt limit t to 3

With this one the problem is we have ln (0). What is the value of ln(0)?



Thannks fo r any help

 

[BobG]

I have couple of problems that I am stuck with.
Hi,
Integration (tan(2x) dx limit pi/4 to 3 pi/4
When I intergrate this, I get ln [1/(sqrt(cos2x))]
So when put the limits, we have zero in the denominator. What do I do?
---------------------------------------------------------------------
The other problem is,
integration (1/(t-3) dt limit t to 3
With this one the problem is we have ln (0). What is the value of ln(0)?
Thannks fo r any help

I think you made a slight mistake there. Instead of getting
ln (cos 2x)^{-(1/2)}}

you should have gotten:
-\frac{1}{2} ln (cos 2x)
Or at least you should have realized the first is actually the same as the second.

By the way, you really didn't even have to finish integrating to find out the answer was zero. If you use u-substitution and change your limits of integration, you never have to substitute the original function back in. In other words:
u=cos 2x \vert_{\frac{\pi}{4}=0}^{\frac{3 \pi}{4}=0}

Since your new lower and upper bounds are equal to each other, you know your result will be zero. If you did continue, you would get to:
du=-2 sin 2x dx
-\frac{1}{2} du = sin 2x dx

Substituting back into your original equation,
-\frac{1}{2} \int_{0}^{0} \frac{1}{u} du
-\frac{1}{2} ln u \vert_{0}^{0}

 

[Gamma]

Thank you,

I understand your point.

But there is nothing wrong in writing

-\frac{1}{2} ln (cos 2x) = ln (cos 2x)^{-(1/2)}right?

By the way is ln(0) defined?

 

[BobG]

Thank you,

I understand your point.

But there is nothing wrong in writing

-\frac{1}{2} ln (cos 2x) = ln (cos 2x)^{-(1/2)}


right?

By the way is ln(0) defined?

Technically, no. I was thinking there had to be a way to avoid the problem, since I know the answer is 0. You just have to rely on the idea that the upper limit and lower limit are the same.

 

[HallsofIvy]

The answer to your last question, "is ln(0) define" is "no". ln(x) is defined only for positive real numbers. (If you extend to complex numbers, ln(x) is defined for for negative x but still not for 0.)

The limit if ln(x), as x goes to 0, is negative infinity.

 

[shmoe]

Be very careful with your first question. When you have an integral that is 'improper' at both endpoints you have to allow your limits to approach these endpoints independantly (unless you are explicitly talking about a Cauchy Principle Value).