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Integration question |
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| Jan28-05, 07:39 PM | #1 |
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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 |
| 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 |
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the second one is 1-sinx^2 not 1+sin^2...
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| Jan28-05, 09:48 PM | #4 |
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Integration questionThanks. |
| Jan28-05, 09:52 PM | #5 |
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Try [tex] x=u-\pi [/tex]
Daniel. |
| Jan28-05, 10:23 PM | #6 |
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with x=u-pi x'=1 (if pi is a constant) sorry i still dont see it lol |
| Jan29-05, 12:30 AM | #7 |
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me neither...
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| 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 |
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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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