## 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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 Blog Entries: 9 Recognitions: Homework Help Science Advisor Is that what u have to show: $$\int_{\pi+\frac{\pi}{6}}^{\pi+\frac{3\pi}{6}} (1+2\sin^{2}x)^{2} dx = \int_{\frac{\pi}{6}}^{\frac{3\pi}{6}} (1+2\sin^{2}x)^{2} dx$$ ???If,so make a substitution...An obvious one. Daniel.
 the second one is 1-sinx^2 not 1+sin^2...

## Integration question

 Quote by dextercioby Is that what u have to show: $$\int_{\pi+\frac{\pi}{6}}^{\pi+\frac{3\pi}{6}} (1+2\sin^{2}x)^{2} dx = \int_{\frac{\pi}{6}}^{\frac{3\pi}{6}} (1+2\sin^{2}x)^{2} dx$$ ???If,so make a substitution...An obvious one. Daniel.
Hey Daniel could you please show me how that'd be solved? im just curious. I have an idea but I dont see it.

Thanks.

 Blog Entries: 9 Recognitions: Homework Help Science Advisor Try $$x=u-\pi$$ Daniel.

 Quote by dextercioby 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 dont see it lol

 me neither...
 Blog Entries: 9 Recognitions: Homework Help Science Advisor $$dx=du$$ 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.
 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.
 Recognitions: Gold Member 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.