Integrate $\int_0^{\pi/6}\arcsin\Bigl(\sqrt{\cos(3\psi)\cos\psi}\Bigr)\,d\psi.$

Question

Question how to evaluate

$$\int_0^{\pi/6}\arcsin\Bigl(\sqrt{\cos(3\psi)\cos\psi}\Bigr)\,d\psi.$$

Attempt:

Let the integral be $I$. We have $$I = \int_0^{\pi/6}\arcsin\Bigl(\sqrt{\cos(3\psi)\cos\psi}\Bigr)\,d\psi.$$ First, we simplify the term inside the square root. Using the triple angle identity for cosine, $\cos(3\psi) = 4\cos^3\psi - 3\cos\psi$,(In order to simplify the integral) we get $$ \cos(3\psi)\cos\psi = (4\cos^3\psi - 3\cos\psi)\cos\psi = \cos^2\psi(4\cos^2\psi - 3). $$

The integral becomes $$ I = \int_0^{\pi/6}\arcsin\Bigl(\cos\psi\sqrt{4\cos^2\psi - 3}\Bigr)\,d\psi. $$

Alternatively, one can make the substitution $x=2\psi$, so $d\psi = dx/2$. The limits become $0$ to $\pi/3$. $$ I = \frac{1}{2} \int_0^{\pi/3} \arcsin\left(\sqrt{\cos(3x/2)\cos(x/2)}\right) dx. $$

Answer 1

With $\cos^{-1} t = \frac12\cos^{-1}(2t^2-1)$ \begin{align} &\int_0^{\pi/6}\sin^{-1}\sqrt{\cos3x\cos x}\ dx\\ = &\int_0^{\pi/6}\cos^{-1}\left(\sin x\sqrt{1+4\cos^2 x}\right)\ dx\\ = &\ \frac12 \int_0^{\pi/6}\cos^{-1}\left[2\sin^2x(1+4\cos^2x)-1\right]\ \overset{2x\to x}{dx}\\ = &\ \frac14\int_0^{\pi/3}\cos^{-1}\left(2\sin^2x-\cos x\right)\ {dx}=\frac{\pi^2}{20} \end{align} where $\int_0^{\pi/3}\cos^{-1}\left(2\sin^2x-\cos x\right)\ {dx}=\frac{\pi^2}5$.

Answer 2

$$I=\int_0^{\frac \pi 6}\arcsin\Bigl(\sqrt{\cos(3x)\cos(x)}\Bigr)\,dx$$ If we use the series expansion of the arc sine function, we should have to compute $$\sum_{n=0}^\infty\frac{ (2 n)!}{4^n\,(2 n+1) (n!)^2}\int_0^{\frac \pi 6}\Big(\cos (3x) \cos ( x)\Big)^{\frac{2 n+1}{2} }\,dx$$ $$J_n=\int_0^{\frac \pi 6}\Big(\cos (3x) \cos ( x)\Big)^{\frac{2 n+1}{2} }\,dx$$ Given by a CAS, the corresponding antiderivative is given in terms of the Appell hypergeometric function of two variables and $$J_n=-\frac{\pi ^{3/2}\, 4^n}{\Gamma \left(-2 n-\frac{1}{2}\right) \Gamma (2 n+2)}\, _2F_1\left(-2 n-1,-n-\frac{1}{2};-2 n-\frac{1}{2};\frac{3}{4}\right)$$ which are all positive.

So, the summand becomes $$a_n=-\frac{\pi ^{3/2}\, _2F_1\left(-2 n-1,-n-\frac{1}{2};-2 n-\frac{1}{2};\frac{3}{4}\right)}{(2 n+1)^2 \, \Gamma \left(-2 n-\frac{1}{2}\right)\, \Gamma (n+1)\, \Gamma (n+1)}$$

For large $n$ $$\frac {a_{n+1}}{a_n}=1-\frac 2n+\frac 4{n^2}+O\left(\frac{1}{n^3}\right)$$ which is not a very good sign for the convergence.

$$\left( \begin{array}{cc} p & \sum_{n=0}^p \\ 10 & 0.483361 \\ 20 & 0.488167 \\ 30 & 0.489879 \\ 40 & 0.490756 \\ 50 & 0.491290 \\ 60 & 0.491649 \\ 70 & 0.491906 \\ 80 & 0.492101 \\ 90 & 0.492252 \\ 100 & 0.492374 \\ 200 & 0.492924 \\ 300 & 0.493109 \\ 400 & 0.493201 \\ 500 & 0.493257 \\ 600 & 0.493294 \\ 700 & 0.493321 \\ 800 & 0.493341 \\ 900 & 0.493356 \\ 1000 & 0.493369 \\ \cdots & \cdots \\ \infty & 0.493480\\ \end{array} \right)$$

Using Wolfram Alpha, this is $\frac {\pi^2}{20}$.

I shall wait for a proof of this result.