How to solve the integral $I_n=\int_{0}^{1}x^n\sqrt{1-x}$?

Question

Let $I_n=\int_{0}^{1}x^n\sqrt{1-x}$. I need to find a recurrence relation, but I do not see any way to do it. Can you help me?

Answer 1

By integration by parts, for $n \geq 1$

\begin{align*} I_n &= \left[ -\frac{2}{3}x^n (1-x)^{3/2} \right]_{0}^{1} + \frac{2n}{3} \int_{0}^{1} x^{n-1}(1-x)^{3/2} \, dx \\ &= \frac{2n}{3} \int_{0}^{1} x^{n-1}(1-x)\sqrt{1-x} \, dx \\ &= \frac{2n}{3}(I_{n-1} - I_n). \end{align*}

Solving this in terms of $I_n$, we obtain

$$ I_n = \frac{2n}{2n+3} I_{n-1}.\tag{*} $$

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Remark. Using the gamma function, it follows from the beta function identity that

$$ I_n = B\left( n+1, \tfrac{3}{2} \right) = \frac{n!(\frac{1}{2})!}{(n+\frac{3}{2})!}. $$

Using this we also confirm that $\text{(*)}$ holds.

Answer 2

Observe that $$\begin{align} I_n &=\int_0^1 x^n\sqrt{1-x}dx\\ &=\int_0^1 x^n(1-x)\frac{dx}{\sqrt{1-x}}\\ &=\int_0^1 x^n\frac{dx}{\sqrt{1-x}}-\int_0^1 x^{n+1}\frac{dx}{\sqrt{1-x}}\\ &= J_n-J_{n+1} \end{align}$$ ... if we define $J_n$ as follows: $$J_n := \int_0^1 x^n\frac{dx}{\sqrt{1-x}}$$ It follows by IBP that $$\begin{align} J_n &=\int_0^1 x^n\frac{dx}{\sqrt{1-x}}\\ &=\bigg[-2x^n\sqrt{1-x}\bigg]_0^1 +2n\int_0^1 x^{n-1}\frac{dx}{\sqrt{1-x}}\\ &=2n\int_0^1 x^{n-1}\sqrt{1-x}dx\\ &=2nI_{n-1}\\ \end{align}$$ So we have $$I_n=J_n-J_{n+1}$$ and $$J_n=2nI_{n-1}$$ Thus, we have $$I_n=2nI_{n-1}-2(n+1)I_n$$ $$(2n+3)I_n=2nI_{n-1}$$ and so $$\color{green}{I_n=\frac{2n}{2n+3}I_{n-1}}$$

Answer 3

Let $\sqrt{1-x}=\cos t, x=\sin^2t\implies dx=2\sin t\cos t\ dt$

$$I_n=\int_0^1x^n\sqrt{1-x}dx=\int_0^{\pi/2}\sin^{2n}t\cos t(2\sin t\cos t\ dt)$$

$$=2\int\cos^2t\sin^{2n+1}t\ dt=B(x,y)$$

where $2x-1=2\iff x=?,2y-1=2n+1\iff$

See Beta Function