Integral $\int\sqrt{\sin2x}\operatorname d\!x$

Question

I tried all substitutions but failed. I need assistance to evaluate that indefinite integral.

$\int\sqrt{\sin2x}\operatorname d\!x$

Answer 1

As said by 5xum, there isn't a simple form for this antiderivative. May be, you could use an infinite Taylor expansion of the integrand such as $$\sqrt{\sin(2x)}=\sqrt{2} \sqrt{x}-\frac{1}{3} \sqrt{2} x^{5/2}+\frac{x^{9/2}}{45 \sqrt{2}}-\frac{x^{13/2}}{189 \sqrt{2}}-\frac{67 x^{17/2}}{56700 \sqrt{2}}-\frac{x^{21/2}}{2772 \sqrt{2}}+O\left(x^{25/2}\right)$$ and integrate.

Answer 2

Let $u=\sqrt{\sin2x}$ ,

Then $x=\dfrac{\sin^{-1}u^2}{2}$

$dx=\dfrac{u}{\sqrt{1-u^4}}du$

$\therefore\int\sqrt{\sin2x}~dx$

$=\int\dfrac{u^2}{\sqrt{1-u^4}}du$

$=\int u^2\sum\limits_{n=0}^\infty\dfrac{(2n)!u^{4n}}{4^n(n!)^2}du$

$=\int\sum\limits_{n=0}^\infty\dfrac{(2n)!u^{4n+2}}{4^n(n!)^2}du$

$=\sum\limits_{n=0}^\infty\dfrac{(2n)!u^{4n+3}}{4^n(n!)^2(4n+3)}+C$

$=\sum\limits_{n=0}^\infty\dfrac{(2n)!\sin^{2n+\frac{3}{2}}2x}{4^n(n!)^2(4n+3)}+C$

Answer 3

Albeit this integral can not by means of Liouville's theorem be resolved in terms of simple function, we can, however, at least, enrich our system of simple function by one more function called elliptic integral of the second kind defined :

$$E(x;k)=\int_0^x\sqrt{\frac{1-k^2 t^2}{1-t^2}} \;\mathrm{d}t$$

This function is standart, sofore well tabelated. Now the system is complete, because, as you can check

$$\int\sqrt{\sin 2x}\;\mathrm{d}x=C-E\left(\frac{\cos x-\sin x}{\sqrt2};2\right)$$