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Evaluate the given integral

$$\int ^{\pi/2}_{0} \sqrt{\sin {2x}} \cdot \sin{x} \cdot dx$$

I am varying various trigonometric manipulation but like reducing it to $\int ^{\pi/2}_{0} \frac{\sin ^2 x}{\sqrt{\tan x}}.dx$ but nothing leads to any fruitful result. Could someone give me some hint?

4 Answers 4

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HINT:

As $\displaystyle\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$

So, if $\int_a^bf(x)\ dx=I,$

$$2I=\int_a^b[f(x)+f(a+b-x)]\ dx$$

$$2I=\int_0^{\pi/2}\sqrt{\sin2x}(\sin x+\cos x)dx$$

As $\displaystyle\int(\sin x+\cos x)dx=-\cos x+\sin x$

Set $-\cos x+\sin x=u\implies\sin2x=1-u^2$

Now use this

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    Always with the best answer. Thanks. – 2017-01-30
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    very nice answer without special functions (+1) – 2017-01-30
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Hint

Use that $\sin(2x) = 2\sin(x)\cos(x)$ and

$$\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} = 2\int^{\pi/2}_0 \cos^{2x-1}(t)\,\sin^{2y-1}(t)\,dt $$

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With $\sin x=u$ and then $u=\sqrt{t}$ we see \begin{align} \int ^{\pi/2}_{0} \sqrt{\sin {2x}} \cdot \sin{x} \cdot dx &=\int_0^1\sqrt{2}u^\frac32(1-u^2)^{-\frac14}du \\ &=\dfrac{\sqrt{2}}{2}\int_0^1t^\frac14(1-t)^{-\frac14}du \\ &=\dfrac{1}{\sqrt{2}}\beta(\dfrac{5}{4},\dfrac{3}{4}) \\ &=\color{blue}{\dfrac{\pi}{4}} \end{align}

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    There is a bug here(you need two square roots for the cosx). I get $\int_0^1\sqrt{2}u^\frac32(1-u^2)^{\frac14}du$ Which after substituting u=x^2 gives $\dfrac{1}{\sqrt{2}}\beta(\dfrac{5}{4},\dfrac{5}{4})$. The answer computed by mathematica and the two answers above is $\pi/4$ or $\dfrac{1}{\sqrt{2}}\beta(\dfrac{5}{4},\dfrac{3}{4})$ – 2017-09-09
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    @user135711 Actually yes. You are right.Thanks – 2017-09-09
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    I made a mistake too, so if you figure it out I'll delete the comment. Otherwise hopefully someone else will see what my bug was, because I'm interested in solving it this way as well. (the $\beta(5/4,5/4)$ I calculated as noted should be $\beta(5/4,3/4)$ – 2017-09-09
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    you write the correct at the end. thanks again. – 2017-09-09
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HINT: write the integral as $$\int_{0}^{\frac {\pi}{2}} (\tan x)^{\frac {3}{2}} \sec^2 x dx $$

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    I think $\cos x$ is missing in your expression. – 2017-01-30
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    I get $\sqrt2(\tan x)^{3/2}\cos^2x$ for the integrand, not $(\tan x)^{3/2}\sec^2x$. – 2017-01-30