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Math Help - Reduction Formula (In terms of x)

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    Reduction Formula (In terms of x)

    Q: Given that I_n = \displaystyle\int^8_0 x^n (8-x)^{\frac{1}{3}}  \, \mathrm{d}x, \ \ n \ge 0.
    (a) Show that I_n = \frac{24n}{3n+4}I_{n-1}, \ \ n \ge 1.
    (b) Hence find the exact value of \displaystyle\int^8_0 x(x+5)(8-x)^{\frac{1}{3}} \, \mathrm{d}x.
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    I've done part (a) and got I_n = 6nI_{n-1} - \frac{3}{4}nI_n but cannot simplify to get it into the form that the question is asking for. Also, I cannot do part (b). Can I have help? Thanks in advance.
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    Quote Originally Posted by Air View Post
    Q: Given that I_n = \displaystyle\int^8_0 x^n (8-x)^{\frac{1}{3}}  \, \mathrm{d}x, \ \ n \ge 0.
    (a) Show that I_n = \frac{24n}{3n+4}I_{n-1}, \ \ n \ge 1.
    (b) Hence find the exact value of \displaystyle\int^8_0 x(x+5)(8-x)^{\frac{1}{3}} \, \mathrm{d}x.
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    I've done part (a) and got I_n = 6nI_{n-1} - \frac{3}{4}nI_n
    Then I_n+\frac 34 n I_n=6n I_{n-1}

    \frac{4+3n}{4} I_n=6n I_{n-1}

    I_n=\frac{24n}{3n+4} I_{n-1}

    You were close to it
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  3. #3
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    Quote Originally Posted by Air View Post
    (b) Hence find the exact value of \displaystyle\int^8_0 x(x+5)(8-x)^{\frac{1}{3}} \, \mathrm{d}x.
    \begin{aligned} \int^8_0 x(x+5)(8-x)^{\frac{1}{3}} \, \mathrm{d}x &=\int_0^8 (x^2+5x)(8-x)^{\frac 13} dx \\<br />
\\<br />
&=\int_0^8 x^2 (8-x)^{\frac 13} dx+\int_0^8 5x(8-x)^{\frac 13} dx \\<br />
\\<br />
&=I_2+5 \cdot I_1 \end{aligned}

    Write I_2 with respect to I_1
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