# Force calculation on a spherical body

#### Morteza

##### New member

$\int \frac{(cos^2(a+b)cos(a)}{x} dx$
Where,
$cos(a) = \frac{r^2+R^2-x^2}{2rR}$and
$cos(b) = \frac{r^2+x^2-R^2}{2rx}$

#### Zermelo

##### Junior Member
Hi there, you should read the guidelines and understand that we’re not allowed to solve your problems, or help you before seeing your work. You need to show us what you did first, and where you’re stuck.
I personally would look up the formula for cos(x + y) if I were you, then the numerator should (eventually) simplify to a polynomial.

#### Subhotosh Khan

##### Super Moderator
Staff member

$\int \frac{(cos^2(a+b)cos(a)}{x} dx$
Where,
$cos(a) = \frac{r^2+R^2-x^2}{2rR}$and
$cos(b) = \frac{r^2+x^2-R^2}{2rx}$
The integrand is an expression with unbalanced parentheses. Please correct it post the correct expression to be integrated.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

#### Morteza

##### New member
Hi there, you should read the guidelines and understand that we’re not allowed to solve your problems, or help you before seeing your work. You need to show us what you did first, and where you’re stuck.
I personally would look up the formula for cos(x + y) if I were you, then the numerator should (eventually) simplify to a polynomial.
Thank you Zermelo for your suggestion

Yes, I did it the same way, but after that, it is a long calculation to get the result and difficult to do that. Actually, I am stuck from the beginning of my work.

#### Subhotosh Khan

##### Super Moderator
Staff member
Thank you Zermelo for your suggestion

Yes, I did it the same way, but after that, it is a long calculation to get the result and difficult to do that. Actually, I am stuck from the beginning of my work.
You have to show us what you did and EXACTLY where you are stuck.

First thing you need to do is to correct your original post.

Then show us how you had used expansion of cos(x + y) and what did you do after that - even if you know that it is incorrect.

#### Morteza

##### New member
I should noted that “r” and “R” are constant.

#### Morteza

##### New member
You have to show us what you did and EXACTLY where you are stuck.

First thing you need to do is to correct your original post.

Then show us how you had used expansion of cos(x + y) and what did you do after that - even if you know that it is incorrect.

#### Morteza

##### New member
$= \int \frac{cos^3(a)cos^2(b)+cos(a)sin^2(a)sin^2(b)-2cos^2(a)cos(b)sin(a)sin(b)} {x} dx$and what would be the next step?

#### topsquark

##### Senior Member
This is the direct way. I usually find it's the more difficult way but looking at this problem, from experience I think (thought, I checked), that there are a lot of cancellations here. But it's going to take some time to work it out.

First, if
[imath]cos(a) = \dfrac{r^2 + R^2 - x^2}{2rR}[/imath] then [imath]sin(a) = \dfrac{\sqrt{(2rR)^2 - (r^2 + R^2 - x^2)^2}}{2rR}[/imath]
and if
[imath]cos(b) = \dfrac{r^2 + x^2 - R^2}{2rx}[/imath] then [imath]sin(b) = \dfrac{\sqrt{(2rx)^2 - (r^2 + x^2 - R^2)^2}}{2rx}[/imath]

Then
[imath]cos(a + b) = cos(a) ~ cos(b) - sin(a) ~ sin(b)[/imath]
$= \left ( \dfrac{r^2 + R^2 - x^2}{2rR} \right ) \cdot \left ( \dfrac{r^2 + x^2 - R^2}{2rx} \right ) - \left ( \dfrac{\sqrt{(2rR)^2 - (r^2 + R^2 - x^2)^2}}{2rR} \right ) \cdot \left ( \dfrac{\sqrt{(2rx)^2 - (r^2 + x^2 - R^2)^2}}{2rx} \right )$
It's a serious mess but the end result of multiplying those square roots is a simple expression: [imath]r^4- 2r^2 (R^2 + x^2) + (R^2 - x^2)^2[/imath].

It's a lot easier from here. Finish out the cos(a + b) calculation, square and multiply by cos(a), then separate everything out in terms of factors of x. Then integrate. It's algebraic busywork but the numerator is just a polynomial in x.

I'll let you fill in the details.

-Dan