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- Thread starter Morteza
- Start date

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- Jan 7, 2021

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I personally would look up the formula for cos(x + y) if I were you, then the numerator should (eventually) simplify to a polynomial.

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The integrand is an expression withPlease find the integration of:

[math]\int \frac{(cos^2(a+b)cos(a)}{x} dx[/math]

Where,

[math]cos(a) = \frac{r^2+R^2-x^2}{2rR}[/math]and

[math]cos(b) = \frac{r^2+x^2-R^2}{2rx}[/math]

Thank you for your help.

Please show us what you have tried and

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

Please share your work/thoughts about this problem

Thank you Zermelo for your suggestion

I personally would look up the formula for cos(x + y) if I were you, then the numerator should (eventually) simplify to a polynomial.

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.

- Joined
- Jun 18, 2007

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- 25,213

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

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.

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.

- Joined
- Aug 27, 2012

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- 1,212

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]

[math] = \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 )[/math]

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