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Integration using reduction formula

Hey I need some help on this question.

I need to integrate the following by parts:

(1-x^2)^n with respect to dx with the limits 0 and 1

And by doing this have to show that the integral is equal to the following:

ImageUploadedByStudent Room1395942286.707285.jpg

I know that proofs aren't allowed but I just need a basic outline in what to do

Thank you much appreciated




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Original post by puddinboy
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So far you know that you need to use integration by parts:

abudvdx dx=uvababvdudx dx\displaystyle \int_{a}^{b} u \cdot \dfrac{\text{d}v}{\text{d}x} \text{ d}x = uv \Big|_{a}^{b} - \int_{a}^{b} v \cdot \dfrac{\text{d}u}{\text{d}x} \text{ d}x

I'll get you started.

 01(1x2)n dx=011(1x2)n dx\displaystyle \because \ \int_{0}^{1} \left( 1 - x^2 \right)^n \text{ d}x = \int_{0}^{1} 1 \cdot \left( 1 - x^2 \right)^n \text{ d}x

You must define your variables as follows:

u=(1x2)nu = \left( 1 - x^2 \right)^n

dvdx=1\dfrac{\text{d}v}{\text{d}x} = 1
(edited 10 years ago)
Reply 2
Original post by Khallil
So far you know that you need to use integration by parts:

abudvdx dx=uvababvdudx dx\displaystyle \int_{a}^{b} u \cdot \dfrac{\text{d}v}{\text{d}x} \text{ d}x = uv \Big|_{a}^{b} - \int_{a}^{b} v \cdot \dfrac{\text{d}u}{\text{d}x} \text{ d}x

I'll get you started.

 01(1x2)n dx=011(1x2)n dx\displaystyle \because \ \int_{0}^{1} \left( 1 - x^2 \right)^n \text{ d}x = \int_{0}^{1} 1 \cdot \left( 1 - x^2 \right)^n \text{ d}x

You must define your variables as follows:

u=(1x2)nu = \left( 1 - x^2 \right)^n

dvdx=1\dfrac{\text{d}v}{\text{d}x} = 1


I did the integral of that it's just I don't know where to go from there


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Original post by puddinboy
I did the integral of that it's just I don't know where to go from there


Posted from TSR Mobile


01(1x2)ndx=[x(1x2)n]01+2n01x2(1x2)n1dx\int_0^1 (1-x^2)^n dx=[x(1-x^2)^n]_0^1+2n\int_0^1 x^2(1-x^2)^{n-1} dx
The first term on the right hand side is zero.
Can you see how to express the integral on the right hand side in terms of In and In1I_n \mathrm{\ and\ }I_{n-1`} ?
Original post by puddinboy
I did the integral of that it's just I don't know where to go from there


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You need to make use of an algebraic "trick" you are unlikely to spot yourself.

x2=1(1x2)x^2=1-(1-x^2)
Reply 5
Original post by brianeverit
01(1x2)ndx=[x(1x2)n]01+2n01x2(1x2)n1dx\int_0^1 (1-x^2)^n dx=[x(1-x^2)^n]_0^1+2n\int_0^1 x^2(1-x^2)^{n-1} dx
The first term on the right hand side is zero.
Can you see how to express the integral on the right hand side in terms of In and In1I_n \mathrm{\ and\ }I_{n-1`} ?


Yeah I think so but I still can't do it


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Original post by puddinboy
Yeah I think so but I still can't do it


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The result follows from incorporating the advice given by Mr M above.
Reply 7
Original post by Indeterminate
The result follows from incorporating the advice given by Mr M above.


I've figured it out


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