Asking For Detailed Integrals Solutions

[Tales Roberto]

Homework Statement

I would like a detailed solution of following Integrals:

$$\int^{z_{2}}_{z_{1}} 1/ \left(s^{2}+z^{2}\right)^{1/2} dz$$

Homework Equations

$$1+tan^{2} \theta = sec^{2} \theta$$

The Attempt at a Solution

$$\int 1/ \left(s^{2}+z^{2}\right)^{1/2} dz = \int 1/ \left[s\left(1+ \left(z/s \right)^{2}\right)^{1/2} \right] dz$$

Let

$$z = s \tan \theta \rightarrow dz = s \sec^{2} \theta d \theta$$
$$\theta = tg^{-1} \left(z/s\right)$$

Then

$$\int^{z_{2}}_{z_{1}} 1/ \left[s\left(1+ \left(z/s \right)^{2}\right)^{1/2} \right] dz = \int^{\theta_{2}}_{\theta_{1}} s \sec^{2} \theta/ \left[s\left(1+tan^{2} \theta \right)^{1/2} \right] d\theta$$

$$\int^{\theta_{2}}_{\theta_{1}} s \sec^{2}\theta / \left[s \sec \theta \right] d\theta = \int^{tg^{-1} \left(z_{2}/s\right)}_{tg^{-1} \left(z_{1}/s\right)} sec \theta d \theta$$

However this not give me the answer i need here (i checked final solution):

$$\int^{z_{1}}_{z_{2}} 1/ \left(s^{2}+z^{2}\right)^{1/2} dz = \left[ln\left(z^{2}+\sqrt{z^{2}+s^{2}\right)}\right]\right|^{z_{2}}_{z_{1}}$$


Please, help!

 

[Tales Roberto]

Another one:

$$\int u/ \left(R^{2}+r^{2}+2Rru\right)^{1/2} du$$

 

[Mute]

Are you sure that's not a correct answer? Just because your solution looks different doesn't mean it's not the same!

For instance, note that the hyperbolic trig function sinh and cosh satisfy

$$1 + \sinh^2x = \cosh^2x$$,

so you could just as well make the substitution $z = s\sinh\theta$. Your answer would then be in terms of an inverse hyperbolic function - which can be expressed in terms of a logarithm. I would suspect that's why your answer doesn't match what you looked up.

Edit: this refers to the integral in the first post.

Also, if you want to do fractions, you use: \frac{numerator}{denominator}
Just be careful with the sqiggly brackets when doing complex fractions, as missing just one of those tends to completely ruin the output (my latex pdf compiler won't even compile anything if that happens)

 

[zcd]

you can simplify s sec^2 theta/ s sec theta into
$$\int^{\theta_{2}}_{\theta_{1}} \sec \theta \right] d\theta$$
which equals ln|sec(theta) + tan(theta)|, and then reverse substitute the z.

 

[Mark44]

Another one:

$$\int 1/ \left(R^{2}+r^{2}+2Rru\right)^{1/2} dz$$

Since you are integrating with respect to z, and assuming that R, r, and u are not functions of z, what you have there is the integral of a constant. It doesn't matter that the constant is complicated by having a radical in the denominator.

In other words,
$$\int \frac{dz}{K} = \frac{1}{K}z + C$$

On the other hand, if this is what you got after doing a substitution, then you didn't go far enough, since you have both u and dz. When you do a substitution, your new integral should not have any of the old variables in it, including the differential.

 

[Tales Roberto]

Sorry, i made a typing mistake here, the integral is:

$$\int u/ \left(R^{2}+r^{2}+2Rru\right)^{1/2} du$$

 

[Tales Roberto]

you can simplify s sec^2 theta/ s sec theta into
$$\int^{\theta_{2}}_{\theta_{1}} \sec \theta \right] d\theta$$
which equals ln|sec(theta) + tan(theta)|, and then reverse substitute the z.

I cannot deal with arctg z in the argument of sec ( ). Its necessary to my purpose find a solution without arctg or sec ( ). However Mute's suggestion would be usefull, i ll try this one and made another attempt.

 

[Mark44]

Sorry, i made a typing mistake here, the integral is:

$$\int u/ \left(R^{2}+r^{2}+2Rru\right)^{1/2} du$$

I would start with this substitution: w = R² + r² + 2Rru. Another one I would try if the first one didn't get me anywhere is w = (R² + r² + 2Rru)^(1/2)