1. ## Integration!

Hi all,

Can someone please help me with the last part of this question? I am lost at how to apply the other parts of the question to this last part..

Thank you so much!

2. ## Re: Integration!

Originally Posted by Tutu
Hi all,

Can someone please help me with the last part of this question? I am lost at how to apply the other parts of the question to this last part..

Thank you so much!
From part ii) you have derived what is called a reduction formula, OP.

Clearly:

\displaystyle \begin{aligned} \int_0^1 \frac{x^n}{\sqrt{1-x^2}} \ dx & = \frac{n-1}{n} \int_0^1 \frac{x^{n-2}}{\sqrt{1-x^2}} \ dx \\ & = \left(\frac{n-1}{n}\right)\cdot\left(\frac{n-3}{n-2}\right)\int_0^1\frac{x^{n-4}}{\sqrt{1-x^2}} \ dx \\ & \vdots \\ & = \prod_{r=0}^{\frac{n-2}{2}} \left(\frac{n-1 -2r}{n - 2r}\right) \int_0^1 \frac{dx}{\sqrt{1-x^2}} \end{aligned}

for even $n$.

Use $n=6$ until you've reduced $\int_0^1 \frac{x^6}{\sqrt{1-x^2}} \ dx$ in terms of $\int_0^1 \frac{dx}{\sqrt{1-x^2}}$.

Edit: in fact, out of interest, we can find a closed form.

Spoiler:
For even $n:$

$\int_0^1 \frac{x^n}{\sqrt{1-x^2}} \ dx = \frac{(n-1)!}{\left(\frac{n-2}{2}\right)!\left(\frac{n}{2}\right)!2^{n-1}} \int_0^1 \frac{dx}{\sqrt{1-x^2}} = \frac{\pi (n-1)!}{\left(\frac{n-2}{2}\right)!\left(\frac{n}{2}\right)!2^n}$