# Integration question

1. Jan 28, 2005

### physicsss

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

2. Jan 28, 2005

### 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.

3. Jan 28, 2005

### physicsss

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

4. Jan 28, 2005

### digink

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.

5. Jan 28, 2005

### dextercioby

Try $$x=u-\pi$$

Daniel.

6. Jan 28, 2005

### digink

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

7. Jan 29, 2005

### physicsss

me neither...

8. Jan 29, 2005

### 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.

9. Jan 29, 2005

### 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.

10. Feb 1, 2005

### 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.

Last edited: Feb 1, 2005