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Integration question

 
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Jan28-05, 07:39 PM   #1
 

Integration question


My text book did the following without giving an explanination. If anyone can fill me in it would be much appreciated.

the integration of (1+2*sin(x)^2)^2 from pi+pi/6 to pi+(3*pi)/6 is equal to
the integration of (1-2*sin(x)^2)^2 from pi/6 to (3*pi)/6.

thanks
 
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Jan28-05, 08:07 PM   #2
 
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Is that what u have to show:
[tex]\int_{\pi+\frac{\pi}{6}}^{\pi+\frac{3\pi}{6}} (1+2\sin^{2}x)^{2} dx =
\int_{\frac{\pi}{6}}^{\frac{3\pi}{6}} (1+2\sin^{2}x)^{2} dx [/tex]

???If,so make a substitution...An obvious one.

Daniel.
 
Jan28-05, 08:31 PM   #3
 
the second one is 1-sinx^2 not 1+sin^2...
 
Jan28-05, 09:48 PM   #4
 

Integration question


Quote by dextercioby
Is that what u have to show:
[tex]\int_{\pi+\frac{\pi}{6}}^{\pi+\frac{3\pi}{6}} (1+2\sin^{2}x)^{2} dx =
\int_{\frac{\pi}{6}}^{\frac{3\pi}{6}} (1+2\sin^{2}x)^{2} dx [/tex]

???If,so make a substitution...An obvious one.

Daniel.
Hey Daniel could you please show me how that'd be solved? im just curious. I have an idea but I dont see it.

Thanks.
 
Jan28-05, 09:52 PM   #5
 
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Try [tex] x=u-\pi [/tex]

Daniel.
 
Jan28-05, 10:23 PM   #6
 
Quote by dextercioby
Try [tex] x=u-\pi [/tex]

Daniel.
Im only in calc2 right now and even that doesn't help me but I guess I will learn more as I finish the rest of my calc courses.

with x=u-pi x'=1 (if pi is a constant)

sorry i still dont see it lol
 
Jan29-05, 12:30 AM   #7
 
me neither...
 
Jan29-05, 06:13 AM   #8
 
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[tex] dx=du [/tex]


Of course,but to wrote everything in terms of "u",u need to make the substitution both in the argument of [itex] \sin [/itex] and in the limits of integration.

Daniel.
 
Jan29-05, 11:33 AM   #9
 
Do you have any trig identity in mind?
Also, are we on the same page? Are we all talking about how (1+2*sin(x)^2)^2 from pi+pi/6 to pi+(3*pi)/6 is equal to (1-2*sin(x)^2)^2 from pi/6 to (3*pi)/6.
 
Feb1-05, 01:14 AM   #10
 
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Looking at the two equations, there is a lot of reason to be suspicious of the integrals being the same.
[tex]1-2sin^2u=cos2u; 1+2sin^2u=2-cos2u[/tex] (This might make integration a little easier.)

[tex]\int_{\pi/6}^{\pi/2}cos^2(2u)du =\pi/6-\sqrt3/16=.415[/tex]

While the other integral is [tex]3/2\pi-15/16\sqrt3 =6.336[/tex]

P.S. It should have been noted that sin(u-Pi) = -sin(u), but for the square, sin^2(u-Pi) = sin^2(u). Thus, as dextercioby suggests, the substitution z=u-Pi, reduces the integral between 7Pi/6 and 3Pi/2 to :

[tex]\int_\frac{\pi}{6}^\frac{\pi}{2}(1+2\sin^2z)^2dz[/tex]

This has a different sign than the other integral.
 
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