# How to integrate $\sqrt{1-\sin 2x}$?

I want to solve the following integral without substitution: $$\int{\sqrt{1-\sin2x}} \space dx$$

I have: $$\int{\sqrt{1-\sin2x}} \space dx = \int{\sqrt{1-2\sin x\cos x}} \space dx = \int{\sqrt{\sin^2x + \cos^2x -2\sin x\cos x}} \space dx$$

but this can be written in two ways: $\int{\sqrt{(\sin x - \cos x)^2}} \space dx$ or $\int{\sqrt{(\cos x - \sin x)^2} \space dx}$
and it seems to be pretty far from what the real result is: http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427ecvsatkj0te&mail=1

Can I get any hints?

EDIT:

Thank you for your answers! So as you all showed, we have:
$\int{\sqrt{1-\sin2x}} \space dx = \cdots = \int{\sqrt{(\sin x - \cos x)^2}} \space dx$
or
$\int{\sqrt{1-\sin2x}} \space dx = \cdots = \int{\sqrt{(\cos x - \sin x)^2} \space dx} = \int{\sqrt{(-(\sin x - \cos x))^2} \space dx}$

combined: $\int{\sqrt{1-\sin2x}} \space dx = \cdots = \int{\sqrt{(\pm(\sin x - \cos x))^2} \space dx}$

so we actually have: $$\int{|\sin x - \cos x|} \space dx$$

I drew myself a trigonometric circle and if I concluded correctly, we have:
1. $\int{\sin x - \cos x} \space dx$ for $x \in [ \frac{\pi}{4} + 2k\pi , \frac{5\pi}{4} + 2k\pi ] , \space k \in \mathbb{Z}$
2. $\int{\cos x - \sin x} \space dx$ for $x \in [ -\frac{3\pi}{4} + 2k\pi , \frac{\pi}{4} + 2k\pi ] , \space k \in \mathbb{Z}$

... which means:
1. $x \in [ \frac{\pi}{4} + 2k\pi , \frac{5\pi}{4} + 2k\pi ] , \space k \in \mathbb{Z}$ :
$$\int{\sqrt{1-\sin2x}} \space dx = \int{\sin x - \cos x} \space dx = -\cos x - \sin x + C$$
2. $x \in [ -\frac{3\pi}{4} + 2k\pi , \frac{\pi}{4} + 2k\pi ] , \space k \in \mathbb{Z}$ : $$\int{\sqrt{1-\sin2x}} \space dx = \int{\cos x - \sin x} \space dx = \sin x + \cos x + C$$

• Both of your ways are actually identical; $\sqrt{a^2} = \sqrt{(-a)^2} = |a|$; they're also essentially identical to Alpha's result. (Note that you can simplify the square root in its final expression just as you simplified it in your integrand) – Steven Stadnicki May 7 '14 at 17:16
• $-\sin \left( 2\,x \right) +1=2\, \left( \cos \left( x+1/4\,\pi \right) \right) ^{2}$ – Graham Hesketh May 7 '14 at 17:18
• The integral over any interval of width $\pi$ is 2, so there is a $2x/\pi$ term to go with the $\sin x$ and $\cos x$ – Empy2 Aug 20 '14 at 12:34
• – Martin Sleziak Jan 4 '17 at 8:30

The result in wolfram is given by : $$\frac{\sqrt{1-\sin(2x)}(\cos x+\sin x)}{\cos x-\sin x}\tag{1}$$ Note your results from simplifying the integration. You want to check that $(1)$ is equal to either$$\int\sqrt{(\sin x-\cos x)^2}\mathrm dx=\int\sin x\mathrm dx-\int\cos x\mathrm dx=-\cos x-\sin x.\tag{2}$$ or $$\int\sqrt{\cos x-\sin x)^2}\mathrm dx=\int\cos x\mathrm dx-\int\sin x\mathrm dx=\sin x+\cos x\tag{3}$$ Notice that $(2)$ and $(3)$ are equal to $(1)$. It is a trigonometric identity. I will show $(2)$ is equal to $(1)$ here. I will leave you to show that $(3)$ is equal to $(1)$.
Proof of $(1)=(2)$: \begin{aligned}&\frac{\sqrt{1-\sin(2x)}(\cos x+\sin x)}{\cos x-\sin x}=-\cos x-\sin x\\&\iff\sqrt{1-\sin(2x)}(\cos x+\sin x)=-(\cos x+\sin x)(\cos x-\sin x)\\&\iff\sqrt{1-\sin(2x)}=-(\cos x-\sin x)\\&\iff \sqrt{1-\sin(2x)}=\sin x-\cos x\\&\iff 1-\sin(2x)=\sin^2 x-2\sin x\cos x+\cos^2 x\\&\iff 1-2\sin(2x)=1-2\sin x\cos x\\&\iff 1-\sin(2x)=1-\sin(2x)\end{aligned} This shows that $$\int\sqrt{1-\sin(2x)}\mathrm dx=-\cos x-\sin x$$even though Wolfram gives you a result that makes you unsure of your own result! Now, just show for yourself that $(1)=(3)$ and you should be convinced that you came up with a much simpler solution than Wolfram.
hint: $1- \sin (2x) = (\sin x - \cos x)^2$
• Good method but one must remember to simplify the integrand to $|\sin x -\cos x|$ and consider both negative and non-negative cases separately. – Deepak Mar 29 '15 at 0:34
You could also use that $$\sin(x)=\cos\left(\frac\pi2-x\right)$$ and $$1-\cos(2y)=2\sin^2(y)$$ And take care of the signs.
Which is of course equivalent to the other variants, since $$\sin\left(\frac\pi4-x\right)=\frac{\sqrt2}2(\cos(x)-\sin(x))$$