Post by: Jimbo123 on February 16, 2020, 05:27:20 pm
Can somebody help me with the question attached. I am having trouble finding the reduction formula part (a). Any help is much appreciated
Post by: DrDusk on February 16, 2020, 06:16:22 pm
Hi All,
Can somebody help me with the question attached. I am having trouble finding the reduction formula part (a). Any help is much appreciated
Post by: RuiAce on February 16, 2020, 06:46:33 pm
I'm fairly sure the difficulty was in everything after the integration by parts. I also don't think this is the best option as the product rule results in an extra integral to keep track of.
\begin{align*}
u &= \left(1-x^5\right)^n\\
u^\prime &= -5x^4n \left(1-x^5\right)^{n-1}\\
v^\prime &= x\\
v &= \frac{x^2}{2}
\end{align*}
\begin{align*}
I_n &=\int_0^1 x\left(1-x^5\right)^n\,dx \\
&= \left[ \frac{x^2}{2} \left(1-x^5\right)^n \right]_0^1 - \int_0^1 \frac{x^2}{2} \cdot -5x^4n\left(1-x^5\right)^{n-1}\,dx\\
&= -\frac{5n}{2} \int_0^1 x \left(-x^5\right) \left(1-x^5\right)^{n-1}\,dx\\
&= -\frac{5n-1}{2} \int_0^1 x\left[ (1-x^5) - 1 \right] \left(1-x^5\right)^{n-1}\,dx\\
&= -\frac{5n}{2} \int_0^1 x\left(1-x^5\right)^n - x\left(1-x^5\right)^{n-1}\,dx\\
&= -\frac{5n}{2} \int_0^1 x\left(1-x^5\right)^n\,dx + \frac{5n}2 \int_0^1 \left(1-x^5\right)^{n-1}\,dx\\
&= -\frac{5n}{2} I_n + \frac{5n)}2 I_{n-1}\\
\implies \frac{5n+2}{2}I_n &= \frac{5n}{2} I_{n-1}\\
I_n &= \frac{5n}{5n+2}I_{n-1}
\end{align*}