recurrence formula (1 Viewer)

tutor01

Member
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

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

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

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

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

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

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

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

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

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

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tutor01

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

Mechanic

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

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

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

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

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

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

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

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

$\bg_white \therefore nI_n - a^2(n - 1)I_{n - 1} = x^{n - 1} \sqrt{x^2 - a^2}$
the 2nd last line the subscript for a^2(n-1) is n-2 but in your last line it is n-1,

Carrotsticks

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

Clearly just a typo.

nightweaver066

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

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

asianese

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