Integration questions for my astoronomy class

Click For Summary
SUMMARY

This discussion focuses on integration techniques relevant to astronomy, specifically integrating the equation dt = da / (H_0 * (Ω_{m,0}/a + a² * Ω_{Λ,0})^(1/2)). The user seeks clarification on how to perform integration from a = 0 to a = 1/(1+z). A suggested method involves a variable substitution using tan(v) to simplify the integration process. The conversation highlights the importance of understanding variable changes in integration for solving complex equations in astrophysics.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques
  • Familiarity with cosmological parameters such as Hubble constant (H_0), matter density (Ω_{m,0}), and dark energy density (Ω_{Λ,0})
  • Knowledge of trigonometric identities, particularly secant and tangent functions
  • Experience with variable substitution methods in integration
NEXT STEPS
  • Study integration techniques in calculus, focusing on variable substitution
  • Learn about cosmological models and their parameters, including Hubble's Law
  • Explore trigonometric identities and their applications in calculus
  • Practice integrating complex functions relevant to astrophysics
USEFUL FOR

Students in astronomy or astrophysics courses, educators teaching calculus and its applications in science, and anyone interested in advanced integration techniques related to cosmological equations.

b_o3
Messages
52
Reaction score
0
Hi, I have to do a lot of integration questions for my astoronomy class but I'm not really sure how to! for example how do u integrate from a number let's say a = 0, to a = 1/(1+z)... and dt = \frac{da}{H_0 \left(\frac{\Omega_{m,0}}{a} + a^2 \Omega_{\Lambda,0}\right)^{\frac{1}{2}}}

I've done a different kind of simple integration but i have no idea how this works,. thanks!
 
Physics news on Phys.org
dt = \frac{da}{H_0 \left(\frac{\Omega_{m,0}}{a} + a^2 \Omega_{\Lambda,0}\right)^{\frac{1}{2}}}

Is this what you mean ?
 
To integrate \frac{du}{(a^2 + b^2 u^2)^{-\frac{1}{2}}}
make the variable change
\frac{bu}{a} = tan v
and remember that
1 + tan^2 v = sec^2 v
 
yest mentz its that one
 

Similar threads

  • · Replies 18 ·
Replies
18
Views
1K
Replies
6
Views
3K
Replies
6
Views
1K
Replies
11
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
64
Views
5K
  • · Replies 9 ·
Replies
9
Views
952
  • · Replies 3 ·
Replies
3
Views
700