Bingxilin John Xena
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plz help
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Reduction q (1 Viewer)
- Thread starter Bingxilin John Xena
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plz help
Hughmaster
Hugh Entwistle
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Use integration by parts with v' = 1. You will get the following calculation:
Hope that helps.
Hope that helps.
ivanradoszyce
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The solution in Latex:
^n dx$}
\\[5mm]
\text{Using integration by parts } \\[5mm]
u &= (3 - x^2)^n \hspace{60pt} v' = 1 \\[5mm]
u' &= n(3 - x^2)^{n-1} \cdot -2x \hspace{20pt} v = x \\[5mm]
I_n &= uv' - \int vu' dx \\[5mm]
I_n &= \bigg[x(3 - x^2)^n\bigg]_0^{\sqrt{3}} - \int_0^{\sqrt{3}} (-2x^2)n(3 - x^2)^{n-1} dx \\[5mm]
I_n &= \bigg[\sqrt{3}(3 - (\sqrt{3})^2)^n - 0 \bigg] + 2n \int_0^{\sqrt{3}} x^2(3 - x^2)^{n-1} dx \\[5mm]
I_n &= 2n \int_0^{\sqrt{3}} x^2(3 - x^2)^{n-1} dx \\[5mm]
\text{This is the tricky bit, let $x^2 = 3 - (3 - x^2)$.} \\[5mm]
I_n &= 2n \Bigg[ \int_0^{\sqrt{3}} \bigg[3 - (3 - x^2)\bigg](3 - x^2)^{n-1} dx
\Bigg] \\[5mm]
\text{Expand} \\[5mm]
I_n &= 2n \Bigg[ \int_0^{\sqrt{3}} 3(3 - x^2)^{n-1} - (3 - x^2)^1(3 - x^2)^{n-1} dx
\Bigg] \\[5mm]
I_n &= 6n \int_0^{\sqrt{3}} (3 - x^2)^{n-1} dx - 2n \int_0^{\sqrt{3}} (3 - x^2)^n dx \\[5mm]
\text{However} \\[5mm]
I_n &= \int_0^{\sqrt{3}} (3 - x^2)^n dx \\[5mm]
\end{align*}
)
^{n-1} dx \\[5mm]
\text{Therefore} \\[5mm]
I_n &= 6n I_{n-1} - 2nI_n \\[5mm]
2nI_n + I_n &= 6n I_{n-1} \\[5mm]
I_n(2n + 1) &= 6n I_{n-1} \\[5mm]
I_n &= \frac{6n}{2n + 1} I_{n-1} \hspace{20pt} n\ge 1
\end{align*}
)