# Homework Help: Integration of functions: problem with borders

1. Jan 13, 2006

### 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

2. Jan 13, 2006

### BobG

$$ln (cos 2x)^{-(1/2)}}$$

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,
$$-\frac{1}{2} \int_{0}^{0} \frac{1}{u} du$$
$$-\frac{1}{2} ln u \vert_{0}^{0}$$

Last edited: Jan 13, 2006
3. Jan 13, 2006

### Gamma

Thank you,

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?

Last edited: Jan 13, 2006
4. Jan 13, 2006

### BobG

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.

5. Jan 13, 2006

### 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.

6. Jan 13, 2006

### 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).