Solving Reduction Formula: Step-by-Step Guide

[suspenc3]

Hi, I am 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

 

[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?

 

[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 <br /> <br /> v=-cosx<br /> <br /> integration by parts;<br /> <br /> \int sin^nxdx = -cosxsin^n^-^1x + (n-1) \int sin^n^-^2xcos^2xdx<br /> ...

 

[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!)

 

[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

 

[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.

 

[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

 

[suspenc3]

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

 

[suspenc3]

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

thanks

 

[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.