Integrating \int^a_0 \frac {dz}{\sqrt{ s^2+z^2 }}: A Helpful Guide

[yungman]

I don't get this answer, this is my work:

Find $$\int^a_0 \frac {dz}{\sqrt{ s^2+z^2 }}$$

Let $$tan \;\theta = \frac z s \;\Rightarrow dz = s \;sec^2\;\theta \;d \theta, \; sec \;\theta = \frac {\sqrt{s^2 + z^2}} s$$

$$\int^a_0 \frac {dz}{\sqrt{ s^2+z^2 }} = \int^a_0 sec \;\theta \;d\theta = ln| sec \;\theta + tan \;\theta| = ln \left | \frac {\sqrt{s^2+z^2} + z}{s} \right |^a_0$$

But from the book:

$$\int^a_0 \frac {dz}{\sqrt{ s^2+z^2 }} = ln | \sqrt{s^2+z^2} + z |^a_0$$

Can anyone help?
Thanks

 

[Mark44]

$$ln \left | \frac {\sqrt{s^2+z^2} + z}{s} \right | = ln |\sqrt{s^2 + z^2} + z| - ln|s|$$

The ln|s| term is a constant, so your answer and the book's answer differ by only a constant.

 

[yungman]

$$ln \left | \frac {\sqrt{s^2+z^2} + z}{s} \right | = ln |\sqrt{s^2 + z^2} + z| - ln|s|$$

The ln|s| term is a constant, so your answer and the book's answer differ by only a constant.

Thanks for the quick reply, this is a part of a bigger problem, the book never account for ln|s|. Later it substitude a into the definite integral and gave the answer and ln|s| is never part of it. I think the book made a mistake because this is really a simple straight forward problem.

 

[Mark44]

If you evaluate the antiderivative (in the form shown on the right side in post #3) at z = b and z = a, and subtract them, the ln|s| terms will drop out.

At z = b: ln|sqrt(s² + b²) + b| - ln|s|
At z = a: ln|sqrt(s² + a²) + a| - ln|s|

If you subtract the 2nd line above from the first, you have -ln|s| + ln|s|.

 

[yungman]

If you evaluate the antiderivative (in the form shown on the right side in post #3) at z = b and z = a, and subtract them, the ln|s| terms will drop out.

At z = b: ln|sqrt(s² + b²) + b| - ln|s|
At z = a: ln|sqrt(s² + a²) + a| - ln|s|

If you subtract the 2nd line above from the first, you have -ln|s| + ln|s|.

But in the problem, we only substitude with a only. I simplify the problem, actually a=ct in the real problem where c is the speed of light and t is time. The answer is not a variable anymore so the term ln|s| need to be accounted for. that is the thing I don't understand.

 

[Char. Limit]

Try evaluating it at a and 0. You will get:

$$\left(ln\left|\sqrt{s^2+a^2}+a\right| - ln|s|\right) - \left(ln\left|\sqrt{s^2+0^2}+0\right| - ln|s|\right) = \left(ln\left|\sqrt{s^2+a^2}+a\right| - ln\left|\sqrt{s^2+0^2}+0\right|\right) + \left(ln|s| - ln|s|\right)$$

So as you can see, the ln(s) term, being constant, doesn't matter to the definite integral.

 

[Mark44]

It's possible for two people to work the same integration problem and get two different answers. This is fine, as long as the two answers differ by only a constant. Your answer and the book's answer differ by a constant.

Here's an example. Two antiderivatives of 2x are x² and x² + 7. For each of these, the derivative is 2x.

If I have a definite integral, from, say, 1, to 3, both antiderivatives give the same answer.

$$\int_1^3 2x dx = \left . x^2\right |_1^3 = 9 - 1 = 8$$
$$\int_1^3 2x dx = \left . x^2 + 7\right |_1^3 = (9 + 7) - (1 + 7) = 16 - 8 = 8$$

 

[yungman]

Thanks, I got it. It is just strange that the book not even write it out first and then cancel out later.