- #1
baby_1
- 159
- 15
- TL;DR Summary
- The integral of a 5-th polynomial in the tangent form
Hello,
Please see this part of the article.
I need to obtain the ##\rho (\phi)## value after obtaining the c0 to c5 constants of the ##\sigma (\phi)##. But as you can see after finding the coefficients, solving Eq.(1) could be a demanding job(I wasn't able to calculate the integral of Eq(1) and obtain the ##\rho (\phi)## ). I need to know how to find ##\rho (\phi)## values for different ##\phi## values.
$$\frac{1}{\rho (\phi )}\frac{d\rho (\phi)}{d\phi}= \\
tan(\frac{c_{0}+c_{1}\phi+c_{2}\phi^2+c_{3}\phi^3+c_{4}\phi^4+c_{5}\phi^5}{2})=>
\\ln(\rho (\phi ))=\int (tan(\frac{c_{0}+c_{1}\phi+c_{2}\phi^2+c_{3}\phi^3+c_{4}\phi^4+c_{5}\phi^5}{2})d\phi)$$
Or, I follow the wrong way.
The writers implied that after finding the ##\sigma (\phi)## the ##\rho (\phi)## could find easily.
Please see this part of the article.
I need to obtain the ##\rho (\phi)## value after obtaining the c0 to c5 constants of the ##\sigma (\phi)##. But as you can see after finding the coefficients, solving Eq.(1) could be a demanding job(I wasn't able to calculate the integral of Eq(1) and obtain the ##\rho (\phi)## ). I need to know how to find ##\rho (\phi)## values for different ##\phi## values.
$$\frac{1}{\rho (\phi )}\frac{d\rho (\phi)}{d\phi}= \\
tan(\frac{c_{0}+c_{1}\phi+c_{2}\phi^2+c_{3}\phi^3+c_{4}\phi^4+c_{5}\phi^5}{2})=>
\\ln(\rho (\phi ))=\int (tan(\frac{c_{0}+c_{1}\phi+c_{2}\phi^2+c_{3}\phi^3+c_{4}\phi^4+c_{5}\phi^5}{2})d\phi)$$
Or, I follow the wrong way.
The writers implied that after finding the ##\sigma (\phi)## the ##\rho (\phi)## could find easily.