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The exercise in my textbook states

You are given that $$I_n=\int^1_0{x(1-2x^4)^n}dx$$

Show that $$I_n=\frac{(-1)^n}{4n +2} + \frac{2n}{2n+1}I_{n-1}$$

I have started out by splitting the power $$\int^1_0{x(1-2x^4)^n}dx=\int^1_0{x(1-2x^4)(1-2x^4)^{n-1}}dx$$ $$\int^1_0{x(1-2x^4)^{n-1}}dx + \int^1_0{-2x^4(1-2x^4)^{n-1}}dx$$ $$I_n=I_{n-1} + \int^1_0{-2x^5(1-2x^4)^{n-1}}dx$$

Then setting up the second integral for integration by parts

$$ I_n=I_{n-1} + \frac{1}{2}\int^1_0{x^2*-4x^3(1-2x^4)^{n-1}}dx $$

Differentiating $x^2$ and integrating $-4x^3(1-2x^4)^{n-1}$

$$I_n=I_{n-1} + \frac{1}{2}\left(\left[\frac{x^2(1-2x^4)}{n}\right]^1_0 -\frac{2}{n}\int^1_0{x(1-2x^4)^n}dx\right)$$ Evaluating $\left[\frac{x^2(1-2x^4)}{n}\right]^1_0$ $$I_n=I_{n-1} + \frac{1}{2}\left(\frac{(-1)^n}{n} -\frac{2}{n}I_n \right)$$ Collecting like terms $$I_n=I_{n-1} +\frac{(-1)^n}{2n} -\frac{1}{n}I_n$$

$$I_n=\frac{1}{1+\frac{1}{n}}I_{n-1} + \frac{(-1)^n}{2n(1+\frac{1}{n})}$$ Simplifying the fractions

$$I_n=\frac{n}{1+n}I_{n-1} + \frac{(-1)^n}{2n+2}$$

However this is not equivelant to the expression in the question, have I messed up somewhere or missed something out?

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    $\begingroup$ As a sanity check, one has $I_0=\int_0^1 x\,dx=1/2$ and $I_1=\int_0^1 x(1-2x^4)\,dx=1/2-2/6=1/6$. Then $$\frac{1}{1+1}I_0+\frac{-1}{3}=\frac14-\frac13 = -\frac{1}{12}\neq I_1.$$ By contrast, $$\frac{2}{2+1}I_0+\frac{-1}{6}=\frac13-\frac16 = \frac16=I_1.$$ So their recursion is correct and there must be some error. $\endgroup$ Dec 17, 2020 at 20:30
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    $\begingroup$ I believe your integration by parts step is incorrect, I got: $x^2\frac{\left(1-2x^4\right)^n}{2n}$ $\endgroup$
    – person
    Dec 17, 2020 at 20:30
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    $\begingroup$ You want $-8x^3$, not $-4x^3$ $\endgroup$
    – Empy2
    Dec 17, 2020 at 20:38
  • $\begingroup$ @person and Empy2 Thank you for pointing out that mistake I redid it and got back the valid expression, I'll flag this for completion now. $\endgroup$ Dec 17, 2020 at 20:52

2 Answers 2

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Let $J = \int_{0}^{1}x^{5} (1 - 2x^{4})^{n-1} dx$.

Integration by parts gives: $$I_n=\int^1_0{x(1-2x^4)^n}dx = \frac{(-1)^{n}}{2} + 4nJ$$.

Separating the integral as you have done gives:

$$I_n=\int^1_0{x(1-2x^4)^n}dx =\int^1_0{x(1-2x^4)(1-2x^4)^{n-1}}dx=I_{n-1} -2 J$$

Solving for $J$ in the first and substitute it into the second gives the desired reduction formula for $I_{n}$.

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$$I_n=\int_0^1x(1-2x^4)^ndx$$ $u=x^2\Rightarrow dx=\frac{du}{2x}$ and so: $$I_n=\frac12\int_0^1(1-2u^2)^ndu$$


$$2I_n=\int_0^1(1+\sqrt{2}u)^n(1-\sqrt{2}u)^ndu$$ If we do IBP: $$a'=(1+\sqrt{2}u)^n\Rightarrow a=\frac{1}{\sqrt{2}(n+1)}(1+\sqrt{2}u)^{n+1}$$ $$b=(1-\sqrt{2}u)^n\Rightarrow b'=-\sqrt{2}n(1-\sqrt{2}u)^{n-1}$$ so: $$2I_n=\frac{1}{\sqrt{2}(n+1)}\left[(1-2u^2)^n(1+\sqrt{2}u)\right]_0^1+\frac{n}{n+1}\int_0^1(1-2u^2)^{n-1}(1+\sqrt{2}u)^2du$$ $$2I_n=\frac{(-1)^n(1+\sqrt{2})-1}{\sqrt{2}(n+1)}+\frac{n}{n+1}\left(I_{n-1}+2\sqrt{2}\int_0^1u(1-2n^2)^{n-1}du+2\int_0^1u^2(1-2u^2)^{n-1}du\right)$$

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