# One trigonometric integral

1. Apr 23, 2010

### Jimmy84

1. The problem statement, all variables and given/known data
This problem is solved in my book however I dont totally understand it.

find the integral of sin^2 (x)

2. Relevant equations

3. The attempt at a solution[/

the integral of sin^2 (x) = integral of (1-cos2x)/2 dx

= (1/2)x -(1/4)sin2x + C

I dont understand the las step basically how did the integra of (1-cos2x)/2 was changed into 2 integrals?

Was it done this way (1-cos2x)/2 = 1/2 (-cos2x) ?

2. Apr 23, 2010

### Char. Limit

Actually, your last step is wrong... if you factor out the one half, you get this.

$$\int sin^2(x) dx = \int \frac{1-cos(2x)}{2} dx = \int \frac{1}{2} - \frac{cos(2x)}{2} dx$$

Which is solvable. Basically, your quote above...

Should be...

See the difference?

3. Apr 23, 2010

### Stratosphere

You actually could have just factored out the 1/2 like this:
$$\int sin^2(x) dx = \int \frac{1-cos(2x)}{2} dx =-\frac{1}{2} \int{cos(2x)} dx$$

It simplifies the integral.

4. Apr 23, 2010

### Jimmy84

so it is the integral of 1/2 - (cos2x)/2

Now how can I find the integral of - (cos2x)/2 ?

I have been trying to solve that by parts but I had to solve for more than 4 integrals.
I choosed u= cos 2x dv = 1/2 dx du -sin2x dx v = 1/2 x

5. Apr 23, 2010

### Stratosphere

He made a mistake, he added an additional 1/2 into the integral. The simplified integral is just cos(2x).

6. Apr 23, 2010

### Stratosphere

I made a little mistake my self.

It should be, [
$$\int sin^2(x) dx = \int \frac{1-cos(2x)}{2} dx =-\frac{1}{2} \int\frac{1}{2}-{cos(2x)} dx$$

The 1/2 couldn't have been factored out like that, I though it was multiplication. Sorry, if I mess you up.

Now that we have that established, the integral can be done like this.

$$\int\frac{1}{2} dx-\int{cos(2x)} dx$$

7. Apr 23, 2010

### Jimmy84

No problem, im trying to solve for cos2x but I think im doing something wrong the integrals that I solved gave me a messier problem.

Im doing u = cos2x dv = dx du = -2sen2x v=x

the result of that was x cos2x + the integrl of 2x sen2x and the result becomes messier when I try to solve for that. what am I doing wrong?

8. Apr 23, 2010

### Dick

Stratosphere still doesn't have it right. There is a factor of 1/2 in front of the second integral. And to integrate cos(2x) just do the u-substitution u=2x. No need for parts.

9. Apr 23, 2010

### Jimmy84

so it would be (1 - cos2x)/2 = 1/2 - (1/2) cos2x

Thanks a lot...

10. Apr 23, 2010

### Stratosphere

I did it out myself and I got the right answer. I didn't tell him that he had to add a constant of tow into the second integral and then balance it out with a 1/2 in front of it. I though he was suppose to solve it.

11. Apr 23, 2010

### Char. Limit

Um, Stratosphere, I really did have the right answer the whole time... The integral you showed had $$\int cos(2x) dx$$ with no coefficient of 1/2 in front of it. This isn't so. That coefficient needs to be there. If you multiply the inside integral by two, you need to divide the outside by two... To get a coefficient of 1/4 outside of the u-subbed integral.

I know my multiplication rules, and the distributive property needs to be applied. Both integrals are multiplied by 1/2.

12. Apr 23, 2010

### Stratosphere

Well, either way, I got the right answer my self, I don;'t really know how to explain it though.

And sorry about that, I make little stupid mistakes, (it was right when I woke up).