# Integration Help: Solving dt=da to get t

• trv
In summary, the homework statement is trying to solve a separable equation in a and t with an integrating term. I'm lost, and I'm guessing there's substitution involved, but I'm not sure on how to get the arctan in. Any thoughts? If someone can tell me how to get the arctan in, I think I'll be able to give a better go at it. Thanks in advance.

## Homework Statement

Need to integrate

dt=da (8*pi*rho0/3a-1)^-1/2

to get

t=(8*pi*rho0/3)*arctan(1/sqrt(8*pi*rho0/3a-1))-sqrt(a)*sqrt(8*pi*rho0/3-a)

## The Attempt at a Solution

I'm totally lost. I'm guessing there's substitution involved, but unsure on how to get the arctan in. Any thoughts. If someone can tell me how to get the arctan in I think I'll be able to give a better go at it.

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Could you list the entire problem, because this doesn't make sense. You say you need to integrate $4\pi r^2$. The only variable in there is r, does r depend on t, what are you integrating over, what does it mean? The answer has a $\rho_0$ and an a, yet the original function does not. There is a template for a reason. List the full problem, list all variables including their meaning.

hey sorry, i can't seem to get the latex right. Let me try again

need to integrate

dt=da (8*pi*rho0/3a-1)^-1/2

Its a separable equation in a and t. the rhs being integrated wrt a

the solution should be
t=(8*pi*rho0/3)*arctan(1/sqrt(8*pi*rho0/3a-1))-sqrt(a)*sqrt(8*pi*rho0/3-a)

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dt=da (8*pi*rho0/3a-1)^-1

This will not yield an arctan. I still don't really know what you're trying to calculate.

Also what is $A_n$?

:(.

I'm working through my cosmology notes. It's an integration that comes up while solving the Friedmann equations for the matter dominated cosmos with spatial curvature, k=+1.

So it definitely can't be integrated to get arctan?

Edit:Oh sorry i make a mistake. Please check it again. It should be -1/2 instead.

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This is the equation you want to integrate as I interpret it.

$$dt=da \frac{1}{\frac{8 \pi \rho_0}{3a}-1}$$. Is this what you had in mind?

The Friedman equation for a matter dominated universe with k=+1 is given by:

$$\left(\frac{\dot{a}}{a}\right)^2+\frac{c^2}{a^2}=\frac{\Omega_m}{a^3}$$

Is this what you're trying to solve?

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Close but with the square-root sign around, (8*pi*rho0/3a-1)

i.e. ^-1/2 rather than ^-1

and, yes, u've got the right equation, with omega_m=8*pi*rho_0/3, and c^2=1.

Ah the square root, now it finally makes sense yes you will get an arctan in it now. I will have a look at it.

Thanks a lot for ur patience.

To reduce the amount of constants in our integral we're going to integrate:

$$\int \frac{da}{\sqrt{\frac{\Omega}{a}-1}}$$

You can substitute the values for $\Omega$ back into the equation later.

Start with substituting $u=\frac{\Omega}{a}-1$. Note there will be multiple substitutions involved, but let's start here.

This should go to

$-\int \frac{\Omega du}{(u+1)^2\sqrt{u}}$

if I haven't made a mistake. Then I guess integration by parts. I'll give that a go now.

Correct, now make a substitution $x=\sqrt{u}$. The next step is a trigonometric substitution. Remember $\tan^2 x+1=\sec^2 x$.

Oh ok.

$-\int \frac{2\Omega dx}{(x^2+1)^2}$

Is that right?

Still correct. Can you see what kind of trig substitution to use now?

Does it have to be a trig. substitution? I don't know a trig. function that integrates to arctan x :(. I do know 1/(1+x^2) integrates to arctan x, so I can see we are quite close however.

Well at first sight you may think you're close, because it looks so much like the derivative on of an arctan, but the square makes quite a difference. Substitute $x=\tan s$ and use the trigonometric identity I listed in one of my earlier posts.

$-\int \frac{2\Omega ds}{sec^3s}$

I think.

Edit: dx was a typo. Changed it to ds.

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You have made an error somewhere, although you're close. Show me what you've done. Also why is there an dx in there and not a ds?

Think I found the mistake. Is it (sec s)^2 rather than (sec s)^3 in the equation above?

Correct. Can you see how to proceed from here on?

I don't unfortunately. Maybe secs^2 to 1+(tan s)^2. Then I'd say replace tan s but this just get us back to before our previous step.

It's pretty easy $\sec x=\frac{1}{\cos x} \Rightarrow \frac{1}{\sec^2 x}=\cos^2x$. You can integrate $\cos^2 x$ by using the double angle formula.

Damn are we anywhere close to complete. I think I'll give up on this for today. Gonna try and work on something else. This seems too long so is unlikely to come up in my exam I'm guessing.

Anyway I hope you'll have a look at the thread tomorow if you have the time. And thanks a lot for your help.

We are very close to completion. After you've integrated cos^2x all that is left to do is back substitute to the original variable a.
It is doubtful that you would be asked to calculate this integral on an exam, it's simply too much work and too little physics. Either way since you're so close I do recommend you finish it.

I'll make sure I have a look at this thread tomorrow.