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:
$\displaystyle \displaystyle \begin{aligned} \int_0^1 \frac{x^n}{\sqrt{1-x^2}} \ dx & = \frac{n-1}{n} \int_0^1 \frac{x^{n-2}}{\sqrt{1-x^2}} \ dx \\ & = \left(\frac{n-1}{n}\right)\cdot\left(\frac{n-3}{n-2}\right)\int_0^1\frac{x^{n-4}}{\sqrt{1-x^2}} \ dx \\ & \vdots \\ & = \prod_{r=0}^{\frac{n-2}{2}} \left(\frac{n-1 -2r}{n - 2r}\right) \int_0^1 \frac{dx}{\sqrt{1-x^2}} \end{aligned}$
for even $\displaystyle n$.
Use $\displaystyle n=6$ until you've reduced $\displaystyle \int_0^1 \frac{x^6}{\sqrt{1-x^2}} \ dx$ in terms of $\displaystyle \int_0^1 \frac{dx}{\sqrt{1-x^2}} $.
Edit: in fact, out of interest, we can find a closed form.
Spoiler: