Integration and hyperbolic function problem According to my cosmology book:$$a\dot{a}^{2}=\frac{\varLambda}{3}a^{3}+K,$$where $a$ is the scale factor, the dot indicates derivative wrt time and $\varLambda$ and $K$ are constants. The author then says “Introducing a new variable $x$ by $a^{3}=x^{2}$ and integrating once more with the initial condition $a\left(0\right)=0$ we obtain”$$a^{3}=\frac{3K}{\varLambda}\sinh^{2}\left(\frac{t}{t_{\varLambda}}\right),$$where $t_{\varLambda}=\frac{2}{\sqrt{3\varLambda}}$. I've tried (with difficulty) integrating the first equation using WolframAlpha but end up with nothing like the second equation. Any suggestions or advice, please?
 A: $$a\dot{a}^{2}=\frac{\varLambda}{3}a^{3}+K,$$
$$x^2=a^3 \implies 3a^2 \dot a =2\dot x x \implies 9a^4 \dot a ^2=4x^2\dot x ^2 $$
$$\implies 9a^3(a\dot a^2)=4x^2\dot x^2 \implies (a\dot a ^2)=\frac {4\dot x^2}{9}$$
The equation becomes
$$a\dot{a}^{2}=\frac{\varLambda}{3}a^{3}+K,$$
$$\frac {4\dot x^2}{9}=\frac{\varLambda}{3}x^2+K$$ 
$$2\dot x=\pm\sqrt {3\varLambda x^2+9K}$$
I integrate for the positive equation...
$$ \int \frac {dx}{\sqrt { x^2+3K/\varLambda }}= \frac{\sqrt {3\varLambda }}{2}\int dt$$
Use a $\sinh(..)$ substitution for the integral and square the result to get $a^3$
$$s^2=3K/\varLambda $$
Substitute $x=s\sinh(m) \implies dx=s\cosh(m)$...for the positive equation we get
$$ \int \frac {dx}{\sqrt { x^2+s^2 }}= \frac{\sqrt {3\varLambda }}{2}\int dt$$
$$ m= \frac{\sqrt {3\varLambda }}{2}t+C$$
$$ arcsinh(x/s)= \frac{\sqrt {3\varLambda }}{2}t+C$$
$$ x= s\sinh(\frac{\sqrt {3\varLambda }}{2}t+C)$$
for the initial condition given $C=0$
$$\boxed{ a^3(t)= \frac {3K}{\varLambda}\sinh^2(\frac{\sqrt {3\varLambda }}{2}t)}$$

Edit for the integral substitute $x=s \sinh(m)$
$$I=\int \frac {dx}{\sqrt { x^2+s^2 }}= \int \frac {s\cosh(m)dm}{\sqrt { s^2\sinh^2(m)+s^2 }}$$
$$I= \int \frac {s\cosh(m)dm}{\sqrt { s^2(\sinh^2(m)+1) }}$$
since you have the equality for hyperbolic functions $\cosh^2(m)-\sinh^2(m)=1$
$$I= \int \frac {s\cosh(m)dm}{\sqrt { s^2\cosh^2(m) }}$$
$$I= \int dm=m+K$$
$$I=arcsinh(x/s)+K$$
A: Use that $2x\dot x=3a^2\dot a$ so that $4x^2\dot x^2=9a^4\dot a^2=9x^2a\dot a^2$ and then 
$$
4\dot x^2=9a\dot a^2=3Λx^2+9K
$$
which you can now solve with standard substitutions.
A: Using Mathematica:
sol = DSolve[{a[t] a'[t]^2 == Λ/3*a[t]^3 + K, a[0] == 0}, a[t], t] // Quiet

(*{{a[t] -> -(((-3)^(1/3) K^(1/3) (-Tanh[1/2 Sqrt[3] t Sqrt[Λ]])^(
    2/3))/(-Λ (-1 + Tanh[1/2 Sqrt[3] t Sqrt[Λ]]^2))^(1/3))},
  {a[t] -> (3^(1/3) K^(1/3) (-Tanh[1/2 Sqrt[3] t Sqrt[Λ]])^(2/3))/(-Λ (-1 + 
   Tanh[1/2 Sqrt[3] t Sqrt[Λ]]^2))^(1/3)},
  {a[t] -> ((-1)^(2/3) 3^(1/3) K^(1/3) (-Tanh[1/2 Sqrt[3] t 
   Sqrt[Λ]])^(2/3))/(-Λ (-1 + Tanh[1/2 Sqrt[3] t Sqrt[Λ]]^2))^(1/3)}, 
  {a[t] -> -(((-3)^(1/3) K^(1/3)Tanh[1/2 Sqrt[3] t Sqrt[Λ]]^(2/3))/(-Λ (-1 + 
   Tanh[1/2 Sqrt[3] t Sqrt[Λ]]^2))^(1/3))}, 
  {a[t] -> (3^(1/3) K^(1/3) Tanh[1/2 Sqrt[3] t Sqrt[Λ]]^(2/3))/(-Λ (-1 + 
  Tanh[1/2 Sqrt[3] t Sqrt[Λ]]^2))^(1/3)}, 
  {a[t] -> ((-1)^(2/3) 3^(1/3) K^(1/3)Tanh[1/2 Sqrt[3] t Sqrt[Λ]]^(
   2/3))/(-Λ (-1 + Tanh[1/2 Sqrt[3] t Sqrt[Λ]]^2))^(1/3)} *)

Mathematica found 6 solution.Let's raise the 6 equation to the power of 3,and 
substitution: $\text{t$\Lambda $}=\frac{2}{\sqrt{3} \sqrt{\Lambda }}$
Table[(a[t] /. sol[[n]])^3 // FullSimplify, {n, 1, 6}] /. 1/2 Sqrt[3] Sqrt[Λ] -> 1/tΛ

(*{(3 k Sinh[t/tΛ]^2)/Λ, (3 k Sinh[t/tΛ]^2)/Λ, (3 k Sinh[t/tΛ]^2)/Λ, 
       3 k Sinh[t/tΛ]^2)/Λ, (3 k Sinh[t/tΛ]^2)/Λ, (3 k Sinh[t/tΛ]^2)/Λ} *)

we have 6 identical solutions.One of them:
$$a(t)^3=\frac{3 k \sinh ^2\left(\frac{t}{\text{t$\Lambda $}}\right)}{\Lambda }$$
