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Establish a recurrence relation for a given integral

Define

    \[ I_n = \int_0^1 (1-x^2)^n \, dx. \]

Prove that this integral satisfies the recurrence relation,

    \[ (2n+1)I_n = 2n I_{n-1}. \]

Using this compute I_2, I_3, I_4, and I_5.


Proof. We start by integrating I_n by parts with

    \begin{align*}  u &= (1-x^2)^n & du &= -2xn(1-x^2)^{n-1} \, dx \\  dv &= dx & v &= x.  \end{align*}

Therefore we have

    \begin{align*}  I_n = \int_0^1 (1-x^2)^n \, dx &= \left( uv -  \int v \, du \right) \Bigr \rvert_0^1 \\[9pt]  &= x(1-x^2)^n \Bigr \rvert_0^1 + 2n \int_0^1 x^2 (1-x^2)^{n-1} \, dx \\[9pt]  &= 2n \int_0^1 x^2 (1-x^2)^{n-1} \, dx \\[9pt]  &= 2n \int_0^1 (1-1+x^2)(1-x^2)^{n-1} \, dx \\[9pt]  &= 2n \int_0^1 (1-x^2)^{n-1} \, dx - 2n \int_0^1 (1-x^2)(1-x^2)^{n-1} \, dx \\[9pt]  &= 2n I_{n-1} - 2n I_n \\[9pt] \implies (2n+1)I_n &= 2n I_{n-1}. \qquad \blacksquare \end{align*}

Now, to compute I_2, I_3, I_4, and I_5 using the recurrence relation we first evaluate I_1 to get ourselves started,

    \[ I_1 = \int_0^1 (1-x^2) \, dx = \left(x - \frac{x^3}{3}\right)\Bigr \rvert_0^1 = \frac{2}{3}. \]

Applying the recurrence relation with n = 2 we have

    \[ 5 I_2 = 4 I_1 \quad \implies \quad I_2 = \frac{8}{15}. \]

Then, applying the recurrence relation with n = 2 (and I_2 = \frac{8}{15}) we have

    \[ 7 I_3 = 6 I_2 \quad \implies \quad I_3 = \frac{16}{35}. \]

Next, applying the recurrence relation with n = 3 (and I_3 = \frac{16}{35}) we have

    \[ 9 I_4 = 8 I_3 \quad \implies \quad I_4 = \frac{128}{315}. \]

Finally, applying the recurrence relation with n = 4 (and I_4 =\frac{128}{315}) we have

    \[ 11 I_5 = 10 I_4 \quad \implies \quad I_5 = \frac{256}{693}. \]

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