How to evaluate $\int \sqrt{\sin^{-1}(\sqrt{\phi})} d\phi $?

Question

How do I go about solving the below integral? $$I_1=\int \sqrt{\sin^{-1}(\sqrt{\phi})} d\phi $$

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Background:

I came across the simpler version of this, which required me to evaluate: $$\int\sin^{-1}(\sqrt{\phi})d\phi$$ I got the solution for this: $$\big[\sin^{-1}(\sqrt{\phi})\big(\phi-\frac{1}{2}\big) + \frac{\sqrt{\phi(1-\phi)}}{2}\big] + C$$

But I still don't see this helping me solve the original question (i.e. $I_1$).

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Note:

Wolfram Alpha gives this solution, but I want to know how to arrive at it.

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Edit:

I did a substitution $\phi = \sin^2(\psi)$ to get:

$$\int \sqrt{\psi}\sin(2\psi)d\psi$$

Now I'm stuck...

Answer 1

With $t=\arcsin\sqrt\phi$, we have $\phi=\sin^2t$ and $$I_1=\int \sqrt{\arcsin\sqrt{\phi}}\,d\phi =\int2\sqrt t\sin t\cos t\,dt=\int\sqrt t\sin 2t\,dt.$$ Now, $2t=p^2$, $dt=p\,dp$, and $$I_1=\frac1{\sqrt2}\int p^2\sin p^2\,dp.$$ By parts,

$$I_1=-\frac1{2\sqrt2}\int p\left(\cos p^2\right)'dp=-\frac1{2\sqrt2}p\cos p^2+\frac1{2\sqrt2}\int \cos p^2\,dp,$$

then we need a Fresnel integral.