About the Integral $\int\arcsin\left(\sin^{2}x\right)dx$

Question

$$\int\arcsin\left(\sin^{2}x\right)dx$$ I am not able to find a closed form elementary solution for this, though I have no reason to believe it exists.
But trying out the Definite Integral as follows: $$I=\int_{0}^{\frac{\pi}{2}}\arcsin\left(\sin^{2}x\right)dx$$

Using the Series Expansion for $\arcsin(x)$: $$\arcsin(x)=\sum_{n=0}^{\infty}\frac{\left(2n\right)!}{2^{2n}\left(n!\right)^{2}}\ \frac{x^{2n+1}}{2n+1}$$ $$\arcsin\left(\sin^{2}x\right)=\sum_{n=0}^{\infty}\frac{\left(2n\right)!}{2^{2n}\left(n!\right)^{2}}\ \frac{\left(\sin x\right)^{4n+2}}{2n+1}$$ $$I=\sum_{n=0}^{\infty}\frac{\left(2n\right)!}{2^{2n}\left(n!\right)^{2}\left(2n+1\right)}\ \int_{0}^{\frac{\pi}{2}}\left(\sin x\right)^{4n+2}dx$$

We are aware of the closed form of the Integral: $$\int_{0}^{\frac{\pi}{2}}\left(\sin x\right)^{a}dx=\frac{\sqrt \pi}{2}\frac{\Gamma\left(\frac{a+1}{2}\right)}{\Gamma\left(\frac{a}{2}+1\right)}$$

$$I=\frac{\sqrt \pi}{2}\sum_{n=0}^{\infty}\frac{\left(2n\right)!}{2^{2n}\left(n!\right)^{2}\left(2n+1\right)}\ \frac{\Gamma\left(\frac{4n+3}{2}\right)}{(2n+1)!}$$ $$I=\frac{\sqrt{\pi}}{2}\sum_{n=0}^{\infty}\frac{\Gamma\left(\frac{4n+3}{2}\right)}{2^{2n}\left(n!\right)^{2}\left(2n+1\right)^{2}}\ $$

Wolfram Gives a Closed Form for the Summation in terms of Hypergeometric Function as follows: $$\sum_{n=0}^{\infty}\frac{\Gamma\left(\frac{4n+3}{2}\right)}{2^{2n}\left(n!\right)^{2}\left(2n+1\right)^{2}}\ =\frac{\sqrt{\pi}}{2}\,_{4}F_{3}\left(\frac{1}{2},\frac{1}{2},\frac{3}{4},\frac{5}{4};1,\frac{3}{2},\frac{3}{2};1\right)$$ Hence, $$I=\frac{\pi}{4}\,_{4}F_{3}\left(\frac{1}{2},\frac{1}{2},\frac{3}{4},\frac{5}{4};1,\frac{3}{2},\frac{3}{2};1\right)$$

I have seen some cases where these hypergeometric functions end up in terms of closed form expressions.

Well my question is:

1. Whether the Indefinite Integral has a solution.
2. Whether the Definite Integral has a Closed Form Solution.
3. Whether the Hypergeometric Series has a Closed Form, which will lead us directly to the Definite Integral.

Answer 1

The answer to your second question is YES:

$$\begin{align*} \int_0^{\tfrac\pi2} \arcsin\left(\sin^2x\right) \, dx &= \int_0^1 \frac{\arcsin y^2}{\sqrt{1-y^2}} \, dy \tag1 \\ &= \int_0^1 \left[\int_0^{\sqrt y} \frac{2yz}{\sqrt{1-y^2z^4}}\right] \, \frac{dy}{\sqrt{1-y^2}} \tag2 \\ &= \int_0^1 \left[\int_{z^2}^1 \frac{2y}{\sqrt{1-y^2}\sqrt{1-z^4y^2}} \, dy\right] \, z \, dz \\ &= 2 \int_0^1 \underbrace{\ln \left(z^2 + \sqrt{1+z^4}\right)}_{\equiv \operatorname{arsinh}z^2} \, \frac{dz}z \tag3\\ &= -4 \int_0^1 \frac{z \ln z}{\sqrt{1+z^4}} \, dz \tag4\\ &= - \int_0^1 \frac{\ln w}{\sqrt{1+w^2}} \, dw \tag5 \end{align*}$$

The last expression can be evaluated using this antiderivative, giving a closed form of

$$\boxed{\ln2\ln\left(1+\sqrt2\right)+\frac12\ln^2\left(1+\sqrt2\right) - \frac{\pi^2}{12} + \frac12 \Re \operatorname{Li}_2\left(3+2\sqrt2\right)}$$

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• $(1)$ Substitute $y=\sin x$
• $(2)$ Convert $\arcsin y^2$ to an integral and interchange them
• $(3)$ The antiderivative turns out to be elementary and can be done with an Euler substitution
• $(4)$ Integrate by parts
• $(5)$ Substitute $w=z^2$

Answer 2

Hint: The hypergeometric series admits also a nice series representation. Recalling the recurrence relation $\Gamma(z+1)=z\Gamma(z)$ and the Legendre' duplication formula \begin{align*} \Gamma(z)\Gamma(z+1/2)=2^{1-2z}\sqrt{\pi}\Gamma(2z)\tag{1} \end{align*} we obtain \begin{align*} \color{blue}{\Gamma\left(\frac{4n+3}{2}\right)}&=\Gamma\left(2n+\frac{3}{2}\right)\\ &=\left(2n+\frac{1}{2}\right)\Gamma\left(2n+\frac{1}{2}\right)\\ &=\left(2n+\frac{1}{2}\right)2^{1-4n}\sqrt{\pi}\frac{\Gamma(4n)}{\Gamma(2n)}\tag{$\to$ (1)}\\ &=\frac{1}{2}(4n+1)2^{1-4n}\sqrt{\pi}\frac{(4n-1)!}{(2n-1)!}\\ &\,\,\color{blue}{=\frac{\sqrt{\pi}}{2^{4n+1}}\frac{(4n+1)!}{(2n)!}}\tag{2} \end{align*}

The integral can be written as \begin{align*} \color{blue}{I}&=\frac{\sqrt{\pi}}{2}\sum_{n=0}^{\infty}\frac{\Gamma\left(\frac{4n+3}{2}\right)}{2^{2n}(n!)^2(2n+1)^2}\\ &=\frac{\pi}{4}\sum_{n=0}^{\infty}\frac{1}{2^{6n}}\frac{(4n+1)!}{(2n)!}\,\frac{1}{(n!)^2(2n+1)^2}\tag{$\to$ (2)}\\ &\,\,\color{blue}{=\frac{\pi}{4}\sum_{n=0}^{\infty}\frac{1}{2^{6n}}\binom{4n+1}{2n+1}\binom{2n}{n}\frac{1}{2n+1}} \end{align*}