<s>[INST] 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?
 [/INST] $$a\dot{a}^{2}=\frac{\varLambda}{3}a^{3}+K,$$
$$x^2=a^3 \implies 3a^2 \dot a =2\dot x x \implies 9a^4 \dot a ^2=4x^2\dot x ^2 $$
$$\implies 9a^3(a\dot a^2)=4x^2\dot x^2 \implies (a\dot a ^2)=\frac {4\dot x^2}{9}$$
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 $\cosh^2(m)-\sinh^2(m)=1$
$$I= \int \frac {s\cosh(m)dm}{\sqrt { s^2\cosh^2(m) }}$$
$$I= \int dm=m+K$$
$$I=arcsinh(x/s)+K$$
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