I am currently learning Calculus and I just started studying Integrals. I want to find the general indefinite integral $$\int x\sqrt{\cos x}\,dx$$
I have tried using some substitution such as letting $u=\cos x$ or $u=\sqrt x$, etc. but it seems like none of them is effective here.
I have also tried using online integral calculators to solve this question, but all I got is something like "steps are currently not supported for this problem."
Since even an integral calculator couldn't handle this, I wondered if this question can be solved. Therefore I came back to MathSE to seek help.
Is it solvable? If so, then how? Thanks in advance.
Best Answer
As said, at the price of the "monster" given by @Raffaele in comments, you can compute it.
What you could also do (but at the price of another infinite summation, is to use $$\sqrt{\cos(x)}=t \implies x=\cos ^{-1}\left(t^2\right)\implies dx=-\frac{2 t}{\sqrt{1-t^4}}\,dt$$ to make $$I=-2\int \frac{t^2}{\sqrt{1-t^4}}\, \cos ^{-1}\left(t^2\right)\,dt$$
Now, using $$\frac{t^2}{\sqrt{1-t^4}}=\sum_{n=0}^\infty (-1)^n \binom{-\frac{1}{2}}{n}\, t^{4 n+2}$$ we face the problem of $$J_n=\int t^{4 n+2}\, \cos ^{-1}\left(t^2\right)\,dt$$ which are $$J_n=\frac{t^{4 n+3} \left(2 t^2 \, _2F_1\left(\frac{1}{2},\frac{4n+5}{4};\frac{4n+9}{4};t^4\right)+(4 n+5) \cos ^{-1}\left(t^2\right)\right)}{(4 n+3) (4 n+5)}$$ But, these integrals express also in terms of elliptic integrals of the first kind. Their general form is $$J_n= t \sqrt{1-t^4}P_{n}(t^4)+\frac {t^{4n+3}}{4n+3} \, \cos ^{-1}\left(t^2\right)+a_n\, F\left(\left.\sin ^{-1}(t)\right|-1\right)$$
Concerning the $a_n$'s, they form the sequence $$\left\{\frac{2}{9},\frac{10}{147},\frac{30}{847},\frac{26}{1155},\frac{442}{27797}, \frac{1326}{110561},\frac{11050}{1168101},\frac{320450}{41575743},\cdots\right\}$$ which do not seem to be known in $OEIS$.
However, in a private discussion, @Raymond Manzoni did identify the sequence $$a_n=\frac{2}{3 (4 n+3)}\prod_{k=1}^n \frac{4 k+1}{4 k+3}=\frac{2}{3 (4 n+3)}\frac{\Gamma \left(\frac{7}{4}\right) \Gamma \left(n+\frac{5}{4}\right)}{\Gamma \left(\frac{5}{4}\right) \Gamma \left(n+\frac{7}{4}\right)}$$
I think that the above would be interesting from a computing point of view.
Edit
If we are concerned by $$I=\int_0^{\frac \pi 2} x\sqrt{\cos (x)}\,dx=\frac{4}{15} \, _3F_2\left(1,\frac{5}{4},\frac{5}{4};\frac{7}{4},\frac{9}{4};1\right)$$
Now, using
$$I=-2\sum_{n=0}^\infty (-1)^n \binom{-\frac{1}{2}}{n}\, J_n$$ Integrated between $0$ and $1$, the first two terms of the expressions for $J_n$ are $0$ and we are left with $$J_n=a_n\, K(-1)=\frac{2\, \sqrt{\pi } \,\Gamma \left(n+\frac{5}{4}\right)}{(4 n+3)^2 \,\Gamma \left(n+\frac{3}{4}\right)}$$
$$I=-4 \sqrt{\pi }\sum_{n=0}^\infty (-1)^n \binom{-\frac{1}{2}}{n}\,\frac{ \Gamma \left(n+\frac{5}{4}\right)}{(4 n+3)^2 \,\Gamma \left(n+\frac{3}{4}\right)}$$ $$I=-\frac{\sqrt{\pi }\, \Gamma \left(\frac{1}{4}\right)}{9 \,\Gamma \left(\frac{3}{4}\right)}\, _3F_2\left(\frac{1}{2},\frac{3}{4},\frac{5}{4};\frac{7}{4},\frac{7}{4};1\right )$$
Update
All of the above is so complex that, may be, approximations could be used.
For example, using the $1,400$ years od approximation $$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad \text{for}\qquad -\frac \pi 2 \leq x\leq\frac \pi 2 $$ in this range $$\int x\sqrt{\cos (x)}\,dx \sim \frac{1}{2} \sqrt{\left(\pi ^2-4 x^2\right) \left(x^2+\pi ^2\right)}+\frac{5}{4} \pi ^2 \sin ^{-1}\left(\frac{2 \sqrt{x^2+\pi ^2}}{\sqrt{5} \pi }\right)$$ which, integrated between $0$ and $\frac \pi 2$ would give $$\frac{5 \pi ^3}{8}-\frac{\pi ^2}{4} \left(2+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)\approx 0.785221$$ while the exact value is $\approx 0.784608$.