Stuck on this trig integral

**twilightstr** · February 5th 2009, 07:20 PM

integrate square root of 1-sin2xdx , [o, pie/4]

**Soroban** · February 5th 2009, 08:28 PM

Hello, twilightstr!

$\int^{\frac{\pi}{4}}_0 \sqrt{1-\sin2x}\,dx$

Under the radical, multiply by: $\frac{1+\sin2x}{1+\sin2x}$

. . $\sqrt{\frac{1-\sin2x}{1}\cdot\frac{1+\sin2x}{1+\sin2x}} \;=\;\sqrt{\frac{1-\sin^2\!2x}{1+\sin2x}} \;=\;\sqrt{\frac{\cos^2\!2x}{1 + \sin2x}}\;=\;\frac{\cos2x}{\sqrt{1+\sin2x}}$

And we have: . $\int^{\frac{\pi}{4}}_0 (1 + \sin2x)^{-\frac{1}{2}}(\cos2x\,dx)$

Then let: $u \:=\:1 + \sin2x$

Got it?

**Krizalid** · February 6th 2009, 06:11 AM

$\sqrt{{{\cos }^{2}}2x}=\left| \cos 2x \right|=\cos 2x,$ because this last is positive for $\left[ 0,\frac{\pi }{4} \right].$

**Chop Suey** · February 6th 2009, 08:55 PM

Another way is to recognize that:

$1 - \sin{(2x)} = \sin^2{x} - 2\sin{x}\cos{x} + \cos^2{x} = (\sin{x} - \cos{x})^2$

Also take note that:
$\sqrt{(\sin{x} - \cos{x})^2} = |\sin{x} - \cos{x}| = -(\sin{x}-\cos{x})$ for $\left[ 0, \frac{\pi}{4} \right]$

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