Proving a Formula for a Definite Integral by Induction.

Question

I've been on this for hours and would really appreciate some help, I'm new to induction in general, so sorry if this is a simple question.

Let $ I_{N}=\int_{0}^{1}x^{n}\sqrt{1-x}dx $

Prove that $I_{N}=\frac{4^{n+1}n!(n+1)!}{(2n+3)!}$ for $n\epsilon\mathbb{N}$.

The base case is obviously $n=1$ and can be verified by substituting n into both formulas and integrating by parts. This gives $I_{N}=\frac{4}{15}$ and is pretty simple. For the inductive step I've assumed $$\exists k\epsilon\mathbb{N}:I_{k}=\int_{0}^{1}x^{k}\sqrt{1-x}dx=\frac{4^{k+1}k!(k+1)!}{(2k+3)!}$$ This is my inductive hypothesis. To prove the claim I then need to show that this implies: $$I_{k+1}=\int_{0}^{1}x^{k+1}\sqrt{1-x}dx=\frac{4^{k+2}k!(k+2)!}{(2k+5)!}$$ Which can be simplified to: $$I_{k+1}=\frac{2(k+1)} {2k+5}I_{k}$$ This is where I get stuck. I tried integrating $I_{k+1}$ by parts taking $u=x^{k+1}$ and $v'=\sqrt{1-x}$ which resulted in: $$\int_{0}^{1}x^{k+1}\sqrt{1-x}dx=\frac{-2}{3}x^{k+1}(1-x)^{\frac{3}{2}}+\frac{2}{3}(k+1)\int_{0}^{1}x^{k}(1-x)^{\frac{3}{2}}dx$$ with the first term being evaluated between 0 and 1. I simplified this as follows: $$\int_{0}^{1}x^{k+1}\sqrt{1-x}dx=\frac{2}{3}(k+1)\int_{0}^{1}x^{x}\sqrt{1-x}(1-x)dx$$ as when evaluated between 1 and 0 the first term goes to zero. I recognise that the term under the integral is related to the $I_{k}$ which i have assumed true, but I just really can't figure out how to proceed from here. I've tried integrating by partd again, I've looked for substitutions and I jsut can;t seem to find anything which works. Even a little push in the right direction would be greatly appreciated, thanks.

Answer 1

Note that

$$\begin{align} I_{k+1}-I_k&=\int_0^1 (x^{k+1}-x^k)\sqrt{1-x}\,dx\\\\ &=-\int_0^1 x^k(1-x)^{3/2}\,dx\tag 1 \end{align}$$

Integrating by parts the integral on the right-hand side of $(1)$ with $u=(1-x)^{3/2}$ and $v=x^{k+1}/(k+1)$ reveals that

$$\begin{align} I_{k+1}-I_k&=-\left.\left(\frac{(1-x)^{3/2}x^{k+1}}{k+1}\right)\right|_{0}^{1}-\frac{3}{2(k+1)}\int_0^1 x^{k+1}\sqrt{1-x}\,dx \\\\ &=-\frac{3}{2(k+1)}I_{k+1}\\\\ \left(1+\frac{3}{2(k+1)}\right)I_{k+1}&=I_k\\\\ I_{k+1}&=\frac{1}{1+\frac{3}{2(k+1)}}I_k\\\\ &=\frac{2(k+1)}{2k+5}I_k \tag 2 \end{align}$$

The iteration on the induction hypothesis is

$$\begin{align} I_{k+1}&=\frac{4^{k+2}(k+1)!(k+2)!)}{(2k+5)!}\\\\ &=\frac{2(k+1)}{2k+5}\frac{4^{k+1}k!(k+1)!}{(2k+3)!}\\\\ &=\frac{2(k+1)}{2k+5}I_k \end{align}$$

which is consistent with $(2)$. And we are done!

Answer 2

Let $x=\sin^2t$ for $t\in [0,\pi /2].$ Abbreviating $s$ for $\sin t$ and $c$ for $\cos t ,$ we have $$\int_0^1 x^n\sqrt {1-x}\; dx=\int_{t=0}^{\pi /2} s^{2 n}c\;d(s^2)=\int_{t=0}^{\pi /2} s^{2 n}c\cdot 2 s c\;dt=$$ $$=\int_{t=0}^{\pi /2} 2 s^{2 n+1}c^2\;dt=\int_{t=0}^{\pi /2} 2 s^{2 n+1}(1-s^2)\;dt=$$ $$= 2 F(2 n+1)- 2 F(2 n+3)$$ where $F(m)=\int_0^{\pi /2} \sin^m t\;dt.$

$$\text {We have } F(m+2)=\int_0^{\pi /2}s^{m+2}\;dt= \int_{t=0}^{\pi/2}s^{m+1}\;d(-c).$$ Integrating by parts, noting that $s^{m+1}(-c)=0$ when $t\in \{0,\pi /2\},$ we have $$F(m+2)=\int_0^{\pi /2}c(m+1)s^m(c\;dt) =$$ $$=\int_0^{\pi /2}(m+1)s^m c^2\;dt=\int_0^{\pi /2}(m+1)(s^m-s^{m+2})\;dt=$$ $$=(m+1)(F(m+2)-F(m)).$$ $$\text {Therefore } F(m+2)=\frac {m}{m+1}F(m).$$.......I think you can take it from here.