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?



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

  • $\begingroup$ Always with the best answer. Thanks. $\endgroup$ – Mathgeek Jan 30 '17 at 17:53
  • $\begingroup$ very nice answer without special functions (+1) $\endgroup$ – tired Jan 30 '17 at 21:09


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

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


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}

  • $\begingroup$ 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 $\dfrac{1}{\sqrt{2}}\beta(\dfrac{5}{4},\dfrac{3}{4})$ $\endgroup$ – user5389726598465 Sep 9 '17 at 13:01
  • $\begingroup$ @user135711 Actually yes. You are right.Thanks $\endgroup$ – Nosrati Sep 9 '17 at 13:19
  • $\begingroup$ 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 $\beta(5/4,3/4)$ $\endgroup$ – user5389726598465 Sep 9 '17 at 13:22
  • $\begingroup$ you write the correct at the end. thanks again. $\endgroup$ – Nosrati Sep 9 '17 at 13:25

HINT: write the integral as $$\int_{0}^{\frac {\pi}{2}} (\tan x)^{\frac {3}{2}} \sec^2 x dx $$

  • $\begingroup$ I think $\cos x$ is missing in your expression. $\endgroup$ – Mathgeek Jan 30 '17 at 16:52
  • $\begingroup$ I get $\sqrt2(\tan x)^{3/2}\cos^2x$ for the integrand, not $(\tan x)^{3/2}\sec^2x$. $\endgroup$ – Barry Cipra Jan 30 '17 at 17:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.