yes,i copied correctly. why??MarcusAgrippa said:Ok. Judging from Emmanuel_Euler's Mathematica solution, evaluating the integral you wrote out above is certainly not a freshman exercise. Are you sure that you have copied out your integral correctly? Or is this something that arose in more advanced work?
I believe the question was intended for the OP.Emmanuel_Euler said:yes,i copied correctly. why??
It does usually, but with this one it seems to take forever to load.Arash. said:wolframalpha also doesn't show step by step guide.
The general formula for integrating 1/(1-A*cosx)^3 is:
∫ 1/(1-A*cosx)^3 dx = (1/2)tan(x/2) / (1-A*cosx)^2 + (A/4) ∫ tan(x/2) / (1-A*cosx) dx
You can simplify the integral of 1/(1-A*cosx)^3 by using the trigonometric identity:
tan(x/2) = sin(x) / (1+cosx)
This will reduce the integral to:
∫ sin(x) / (1-A*cosx)^2 dx
The substitution u = tan(x/2) is commonly used to solve the integral of 1/(1-A*cosx)^3. This will transform the integral into a simpler form, as mentioned in the previous answer.
Yes, the integral of 1/(1-A*cosx)^3 can be solved using partial fractions. However, this method may require more complex calculations compared to using the substitution u = tan(x/2).
Yes, when A = 1, the integral of 1/(1-A*cosx)^3 becomes a special case. In this case, the integral can be solved using the substitution u = tan(x/2) and simplifying using the identity:
tan(x/2) = sin(x) / (1+cosx)
The resulting integral will be in the form of ∫ sin(x) / (1-cosx)^2 dx, which can be solved using another substitution.