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?
Best Answer
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 $I_{n}$.