Not able to integrate $\displaystyle\int_{0}^{\pi/2}
\frac{\sin\left(x\right)}
{\left[1 + \,\sqrt{\,\sin\left(2x\right)\,}\,\right]^{2}} \,\mathrm{d}x$
i used the property to change reach
 $\displaystyle 2I =
\int_0^{\pi/2}\frac{\sin\left(x\right) + \cos\left(x\right)}
{\left[1 + \,\sqrt{\,\sin\left(2x\right)\,}\,\right]^2} \,\mathrm{d}x$
now writing $\displaystyle\sin\left(2x\right) = 
1 - \left[\sin\left(x\right) - \cos\left(x\right)\right]^{\, 2}$, and substituting
$\displaystyle\sin\left(x\right) - \cos\left(x\right) = t$,
how to do further ?.
 A: As in the OP, the substitution $y=\pi/2-x$ gives
$$
\begin{aligned}
J&:=
\int_0^{\pi/2} \frac {\sin x}{(1+ \sqrt{\sin 2x})^2} \;dx
\\
&=\int_0^{\pi/2} \frac {\cos x}{(1+ \sqrt{\sin 2x})^2} \;dx
\text{ ... and thus}
\\
&=
\frac 12
\int_0^{\pi/2} \frac {\sin x+\cos x}{(1+ \sqrt{\sin 2x})^2} \;dx\dots
\\
&\qquad\qquad\text{Now formally set $t=\sin x-\cos x$,}
\\
&\qquad\qquad\text{so $dt=\cos x+\sin x$,
$t^2=1-2\sin x\cos x=1-\sin 2x$...}
\\
&=
\frac 12
\int_{-1}^{1}
\frac {dt}{(1+ \sqrt{1-t^2})^2}
=
\int_0^1
\frac {dt}{(1+ \sqrt{1-t^2})^2}
\\
&\qquad\qquad\text{Now use $t=\sin u$}
\\
&=
\int_0^{\pi/2}
\frac {\cos u\; du}{(1+ \cos u)^2}
\\
&\qquad\qquad\text{Now use $v=\tan(u/2)$}
\\
&=
\int_0^1
\frac {\frac{1-v^2}{1+v^2}\cdot\frac 2{1+v^2}\; dv}
{\left(\frac 2{1+v^2}\right)^2}
=
\frac 12
\int_0^1(1-v^2)\; dv
\\
&=\frac 12\left[v-\frac 13 v^3\right]_0^1
=
\frac 12\cdot \frac 23 =\frac 13\ .
\end{aligned}
$$
Computer check, pari/gp:
? intnum( x=0, Pi/2, sin(x) / ( 1+sqrt(sin(2*x)) )^2 )
%1 = 0.33333333333333333333333333333333333333

