Integration Q: 1+2sin^2(x)^2 to 1-2sin^2(x)^2 from pi/6-3pi/6

[physicsss]

My textbook 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

 

[dextercioby]

Is that what u have to show:
\int_{\pi+\frac{\pi}{6}}^{\pi+\frac{3\pi}{6}} (1+2\sin^{2}x)^{2} dx =<br /> \int_{\frac{\pi}{6}}^{\frac{3\pi}{6}} (1+2\sin^{2}x)^{2} dx

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

Daniel.

 

[physicsss]

the second one is 1-sinx^2 not 1+sin^2...

 

[digink]

Is that what u have to show:
\int_{\pi+\frac{\pi}{6}}^{\pi+\frac{3\pi}{6}} (1+2\sin^{2}x)^{2} dx =<br /> \int_{\frac{\pi}{6}}^{\frac{3\pi}{6}} (1+2\sin^{2}x)^{2} dx

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

Daniel.

Hey Daniel could you please show me how that'd be solved? I am just curious. I have an idea but I don't see it.

Thanks.

 

[dextercioby]

Try x=u-\pi

Daniel.

 

[digink]

Try x=u-\pi

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 don't see it lol

 

[physicsss]

me neither...

 

[dextercioby]

dx=du


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

Daniel.

 

[physicsss]

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.

 

[robert Ihnot]

Looking at the two equations, there is a lot of reason to be suspicious of the integrals being the same.
1-2sin^2u=cos2u; 1+2sin^2u=2-cos2u (This might make integration a little easier.)

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

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

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 :

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

This has a different sign than the other integral.