# Reduction Formula

1. Jun 9, 2006

### suspenc3

Hi, Im having trouble understanding this question, I have looked over a few examples, but I'm still confused about the process.

A)Use the reduction formula to show that:

$$\int sin^2xdx = \frac {x}{2} - \frac{sin2x}{4} + C$$

any help would be appreciated

2. Jun 9, 2006

### TD

I assume you're referring to the reduction of the exponent?

Using cos(2x) = cos²x-sin²x combined with cos²x+sin²x = 1, you can derive the following formulas to get rid of a square in cos or sin:

sin²x = (1-cos(2x))/2 and cos²x = (1+cos(2x))/2

Try to verify this yourself.

Now, using the first formula, do you see how the integral was done?

3. Jun 9, 2006

### suspenc3

i am still confused, this is the first question like this I have done. The question says to refer to ex.6...here it is:

$$\int sin^nxdx = -\frac{1}{n}cosxsin^n^-^1x + \frac{n-1}{n} \int sin^n^-^2xdx$$

let: $$u=sin^n^-^1$$

$$dv=sinxdx$$

$$du = (n-1)sin^n^-^xcosxdx [tex] v=-cosx$$

integration by parts;

$$\int sin^nxdx = -cosxsin^n^-^1x + (n-1) \int sin^n^-^2xcos^2xdx$$
..........

4. Jun 9, 2006

### TD

I see, they really mean a reduction formula for the integral (a bit overkill for such an integral, imho).

In that case, compare the formula (your first line) with the problem. It's exactly the same, only n = 2.
So apply the formule with n = 2, no integration by parts is necessary (unless you'd want to prove the reduction formula, but that isn't asked here!)

5. Jun 9, 2006

### suspenc3

$$\int sin^2xdx = \frac {x}{2} - \frac{sin2x}{4} + C$$

so $$u=sin^n^-^1$$...i get that part..and end up with only -sinx

6. Jun 9, 2006

### TD

Are you trying to prove the reduction formula you gave?
I don't understand why you keep coming that this 'u' for a substitution.

I understand the problem as:

Find

$$\int sin^2xdx$$

Using the formula

$$\int sin^nxdx = -\frac{1}{n}cosxsin^n^-^1x + \frac{n-1}{n} \int sin^n^-^2xdx$$

Is that what you're supposed to do? If so, apply this last formula with n = 2.

7. Jun 9, 2006

### suspenc3

I am trying to show$$\int sin^2xdx = \frac {x}{2} - \frac{sin2x}{4} + C$$
using the reduction formula shown in example 6:$$\int sin^nxdx = -\frac{1}{n}cosxsin^n^-^1x + \frac{n-1}{n} \int sin^n^-^2xdx$$

8. Jun 9, 2006

### suspenc3

and in that example they let $$u=sin^n^-^1$$etc...shouldnt i do the same for what i am trying to show?

9. Jun 9, 2006

### suspenc3

i made this too complicated...hahaha so easy..nvm i understand now

thanks

10. Jun 9, 2006

### TD

I think that in example 6, they have proven this formula. In order to do this, they'll have used integration by parts I assume.
What you now have to do (*I think*), is use this formula (not prove it again) on the particular problem.

In ex 6, they've set up a relation between the integral of sin(x)^n and an integral with sin(x)^(n-2), so this formula allows you to reduce the exponent by 2 every time you apply it. Now in your problem, you wish to find the primitive of sin²x, you can use this formula with n = 2.