## Reduction Formula

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

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 Recognitions: Homework Help 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?
 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; $$\int sin^nxdx = -cosxsin^n^-^1x + (n-1) \int sin^n^-^2xcos^2xdx$$ ..........

Recognitions:
Homework Help

## Reduction Formula

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!)

 $$\int sin^2xdx = \frac {x}{2} - \frac{sin2x}{4} + C$$ so $$u=sin^n^-^1$$...i get that part..and end up with only -sinx
 Recognitions: Homework Help 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.
 I am trying to show$$\int sin^2xdx = \frac {x}{2} - \frac{sin2x}{4} + C$$ using the reduction formula shown in example 6:$$\int sin^nxdx = -\frac{1}{n}cosxsin^n^-^1x + \frac{n-1}{n} \int sin^n^-^2xdx$$
 and in that example they let $$u=sin^n^-^1$$etc...shouldnt i do the same for what i am trying to show?
 i made this too complicated...hahaha so easy..nvm i understand now thanks
 Recognitions: Homework Help 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.