# Stuck on this trig integral

• Feb 5th 2009, 07:20 PM
twilightstr
Stuck on this trig integral
integrate square root of 1-sin2xdx , [o, pie/4]
• Feb 5th 2009, 08:28 PM
Soroban
Hello, twilightstr!

Quote:

$\displaystyle \int^{\frac{\pi}{4}}_0 \sqrt{1-\sin2x}\,dx$
Under the radical, multiply by: $\displaystyle \frac{1+\sin2x}{1+\sin2x}$

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

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

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

Got it?

• Feb 6th 2009, 06:11 AM
Krizalid
$\displaystyle \sqrt{{{\cos }^{2}}2x}=\left| \cos 2x \right|=\cos 2x,$ because this last is positive for $\displaystyle \left[ 0,\frac{\pi }{4} \right].$
• Feb 6th 2009, 08:55 PM
Chop Suey
Another way is to recognize that:

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

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