5
$\begingroup$

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

$\endgroup$
  • 3
    $\begingroup$ 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) $\endgroup$ – Steven Stadnicki May 7 '14 at 17:16
  • 2
    $\begingroup$ $-\sin \left( 2\,x \right) +1=2\, \left( \cos \left( x+1/4\,\pi \right) \right) ^{2}$ $\endgroup$ – Graham Hesketh May 7 '14 at 17:18
  • $\begingroup$ The integral over any interval of width $\pi$ is 2, so there is a $2x/\pi$ term to go with the $\sin x$ and $\cos x$ $\endgroup$ – Empy2 Aug 20 '14 at 12:34
  • $\begingroup$ See also Trigonometric integration $ \int \sqrt{1-\sin2x}dx$. Found using Approach0. $\endgroup$ – Martin Sleziak Jan 4 '17 at 8:30
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
7
$\begingroup$

hint: $1- \sin (2x) = (\sin x - \cos x)^2$

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Good method but one must remember to simplify the integrand to $|\sin x -\cos x|$ and consider both negative and non-negative cases separately. $\endgroup$ – Deepak Mar 29 '15 at 0:34
2
$\begingroup$

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 $$ \sin\left(\frac\pi4-x\right)=\frac{\sqrt2}2(\cos(x)-\sin(x)) $$

| cite | improve this answer | |
$\endgroup$

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.