recurrence formula (1 Viewer)

tutor01 · Jun 30, 2012

Given I_n = integral of x^n divided by sqrt(x^2-a^2) dx show that

nI_n - (n-1)a^2I_(n-2)=x^(n-1)sqrt(x^2-a^2).

I can't prove it. Can anyone help. Note I_n means I then subscript n like for recurrence integral questions.

Thanks!

nightweaver066 · Jun 30, 2012

$I_n = \int \frac{x^n}{\sqrt{x^2 - a^2}}dx$

$= \int x^{n - 1} \times \frac{x}{\sqrt{x^2 - a^2}}dx$

$= \int x^{n - 1} d(\sqrt{x^2 - a^2})$

$= x^{n - 1}\sqrt{x^2 - a^2} - (n - 1) \int x^{n - 2} \sqrt{x^2 - a^2}dx$

$= x^{n - 1} \sqrt{x^2 - a^2} - (n - 1) \int x^{n - 2} \sqrt{x^2 - a^2} \times \frac{\sqrt{x^2 - a^2}}{\sqrt{x^2 - a^2}}dx$

$= x^{n - 1} \sqrt{x^2 - a^2} - (n - 1) \int \frac{x^{n - 2}(x^2 - a^2)}{\sqrt{x^2 - a^2}}dx$

$= x^{n - 1} \sqrt{x^2 - a^2} - (n - 1) \int \frac{x^n}{\sqrt{x^2 - a^2}}dx + a^2(n - 1) \int \frac{x^{n - 2}}{\sqrt{x^2 - a^2}}dx$

$= x^{n - 1} \sqrt{x^2 - a^2} - (n - 1)I_n + a^2(n - 1)I_{n - 2}$

$\therefore nI_n - a^2(n - 1)I_{n - 2} = x^{n - 1} \sqrt{x^2 - a^2}$

tutor01 · Jun 30, 2012

Brilliant. I got to line 4 but did not then multiply by the root x^2-a^2 on itself. Thanks a lot.

Mechanic · Jul 9, 2012

nightweaver066 said:
$I_n = \int \frac{x^n}{\sqrt{x^2 - a^2}}dx$

$= \int x^{n - 1} \times \frac{x}{\sqrt{x^2 - a^2}}dx$

$= \int x^{n - 1} d(\sqrt{x^2 - a^2})$

$= x^{n - 1}\sqrt{x^2 - a^2} - (n - 1) \int x^{n - 2} \sqrt{x^2 - a^2}dx$

$= x^{n - 1} \sqrt{x^2 - a^2} - (n - 1) \int x^{n - 2} \sqrt{x^2 - a^2} \times \frac{\sqrt{x^2 - a^2}}{\sqrt{x^2 - a^2}}dx$

$= x^{n - 1} \sqrt{x^2 - a^2} - (n - 1) \int \frac{x^{n - 2}(x^2 - a^2)}{\sqrt{x^2 - a^2}}dx$

$= x^{n - 1} \sqrt{x^2 - a^2} - (n - 1) \int \frac{x^n}{\sqrt{x^2 - a^2}}dx + a^2(n - 1) \int \frac{x^{n - 2}}{\sqrt{x^2 - a^2}}dx$

$= x^{n - 1} \sqrt{x^2 - a^2} - (n - 1)I_n + a^2(n - 1)I_{n - 2}$

$\therefore nI_n - a^2(n - 1)I_{n - 1} = x^{n - 1} \sqrt{x^2 - a^2}$

the 2nd last line the subscript for a^2(n-1) is n-2 but in your last line it is n-1,

your therefore line is incorrect...

Carrotsticks · Jul 9, 2012

Mechanic said:
the 2nd last line the subscript for a^2(n-1) is n-2 but in your last line it is n-1,

your therefore line is incorrect...

Clearly just a typo.

nightweaver066 · Jul 9, 2012

Mechanic said:
the 2nd last line the subscript for a^2(n-1) is n-2 but in your last line it is n-1,

your therefore line is incorrect...

I'll edit it if it's bothering you that much.

asianese · Jul 9, 2012

You can also do this question by splitting $x^n\; $into$\; x^2 x^{n-2}$ and then doing the +a^2-a^2 trick inside the x^2 bracket.

Mechanic · Jul 9, 2012

nightweaver066 said:
I'll edit it if it's bothering you that much.

just saying

recurrence formula (1 Viewer)

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