Reduction Formula Question (1 Viewer)

Lith_30 · Feb 1, 2022

How do I find the reduction formula of $I_n=\int{(\sin^{-1}x)^ndx$ . Thanks

notme123 · Feb 1, 2022

Lith_30 said:
How do I find the reduction formula of $I_n=\int{(\sin^{-1}x)^ndx$ . Thanks

$I_n=\int{(\sin^{-1}x)^ndx$
integrate 1, differentiate $(\sin^{-1}x)^n$
$= x(\sin^{-1}x)^n - n\int{\frac{x(\sin^{-1}x)^{n-1}}{\sqrt{1-x^2}}dx$
integrate $\frac{x}{\sqrt{1-x^2}$ , differentiate $(\sin^{-1}x)^{n-1}$
$= x(\sin^{-1}x)^n - n(-\sqrt{1-x^2}(\sin^{-1}x)^{n-1}+(n-1)\int{\frac{\sqrt{1-x^2}(\sin^{-1}x)^{n-2}}{\sqrt{1-x^2}}dx)$
$= x(\sin^{-1}x)^n - n(-\sqrt{1-x^2}(\sin^{-1}x)^{n-1}+(n-1)\int{(\sin^{-1}x)^{n-2}dx)$
$= x(\sin^{-1}x)^n - n(-\sqrt{1-x^2}(\sin^{-1}x)^{n-1}+(n-1)I_{n-2})$
$= x(\sin^{-1}x)^n + n\sqrt{1-x^2}(\sin^{-1}x)^{n-1}-n(n-1)I_{n-2}$

Lith_30 · Feb 2, 2022

notme123 said:
$I_n=\int{(\sin^{-1}x)^ndx$
integrate 1, differentiate $(\sin^{-1}x)^n$
$= x(\sin^{-1}x)^n - n\int{\frac{x(\sin^{-1}x)^{n-1}}{\sqrt{1-x^2}}dx$
integrate $\frac{x}{\sqrt{1-x^2}$ , differentiate $(\sin^{-1}x)^{n-1}$
$= x(\sin^{-1}x)^n - n(-\sqrt{1-x^2}(\sin^{-1}x)^{n-1}+(n-1)\int{\frac{\sqrt{1-x^2}(\sin^{-1}x)^{n-2}}{\sqrt{1-x^2}}dx)$
$= x(\sin^{-1}x)^n - n(-\sqrt{1-x^2}(\sin^{-1}x)^{n-1}+(n-1)\int{(\sin^{-1}x)^{n-2}dx)$
$= x(\sin^{-1}x)^n - n(-\sqrt{1-x^2}(\sin^{-1}x)^{n-1}+(n-1)I_{n-2})$
$= x(\sin^{-1}x)^n + n\sqrt{1-x^2}(\sin^{-1}x)^{n-1}-n(n-1)I_{n-2}$

Thank you! It was double ibp all along

Reduction Formula Question (1 Viewer)

Lith_30

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notme123

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Lith_30

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