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$\displaystyle\int_{0}^{\pi/2} \frac{\sin\left(x\right)} {\left[1 + \,\sqrt{\,\sin\left(2x\right)\,}\,\right]^{2}} \,\mathrm{d}x$

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 ?.

Felix Marin
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maveric
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1 Answers1

7

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
dan_fulea
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