Question

Show that \(\int_{0}^{1}\left(1-x^{2}\right)^{n} d x=\frac{2^{2 n}(n !)^{2}}{(2 n+1) !}\).

Short answer

Short Answer: The given formula \(\int_{0}^{1}\left(1-x^{2}\right)^{n} dx = \frac{2^{2n}(n!)^{2}}{(2n+1)!}\) can be proven using integration by parts, and induction. We first defined the necessary parameters for integration by parts, then applied the formula to evaluate the integral. After that, we showed the result holds true for the base case, n=0. Using induction, we assumed the result is true for n=k and through simplifications, we showed it also holds true for n=k+1. This confirms that the given formula is correct.

Step-by-step solution

Define parameters for integration by parts

In order to integrate by parts, we will define the following parameters: \(u = \left(1 - x^2\right)^n\), \(dv = dx\). Now, we will differentiate \(u\) to obtain \(du\) and integrate \(dv\) to obtain \(v\): \(du = -2nx^{2n-1} dx\) and \(v = x\)

Apply integration by parts formula

The Integration by parts formula is given by: \(\int udv = uv - \int vdu\). Now we apply this formula with the chosen \(u\), \(v\), \(du\), and \(dv\) we have defined: $$\int_{0}^{1} \left(1-x^2\right)^ndx=\left[x\left(1-x^2\right)^n\right]^1_0 - \int_{0}^{1} x(-2n) x^{2n-1} dx$$ Notice that the term within the bracket evaluates to 0 when x=0 and x=1. Therefore, the first term evaluates to 0, and we're left with the following integral: $$\int_{0}^{1}(1-x^2)^ndx= 2n \int_{0}^{1}x^{2n} dx$$

Evaluate the integral

Now we just need to evaluate the integral: $$2n \int_{0}^{1}x^{2n} dx = 2n \left[\frac{x^{2n+1}}{2n+1}\right]^1_0$$ The right-hand side of the equation evaluates to: $$ 2n \cdot \frac{1^{2n+1}-0^{2n+1}}{2n+1} = 2n \cdot \frac{1^1}{2n+1} = \frac{2n}{2n+1}$$ Now, we know that \(\int_{0}^{1}\left(1-x^{2}\right)^{n} d x=\frac{2n}{2n+1}\).

Use induction to show the result

We will now use induction to show the result. For \(n=0\), $$\int_{0}^{1}\left(1-x^{2}\right)^{0} d x = \int_{0}^{1}1 d x$$ $$\int_{0}^{1}1 d x = x|^1_0 = 1$$ For \(n=0\), the right-hand side of the equation is: $$\frac{2^{2\cdot 0}(0!)^{2}}{(2\cdot0+1)!} = \frac{1^2}{1!} = 1$$ The result holds true for the base case of \(n=0\). Now we assume the result is true for \(n=k\). So, $$\int_{0}^{1}\left(1-x^{2}\right)^{k} d x = \frac{2^{2k}(k!)^{2}}{(2k+1)!}$$ We will show that this holds true for \(n=k+1\) as well. Since we know that \(\int_{0}^{1}\left(1-x^{2}\right)^{n} d x=\frac{2n}{2n+1}\), for \(n=k+1\), we have: $$\int_{0}^{1}\left(1-x^{2}\right)^{k+1} d x = \frac{2(k+1)}{2(k+1)+1}$$ We will assume that this equation holds true and now we will check if the general formula holds for \(n=k+1\). Using the general formula, we can determine the value that should be equal to our integral. $$\frac{2^{2(k+1)}((k+1)!)^{2}}{(2(k+1)+1)!}$$ With some simplifications, this results in: $$\frac{2^{2k+2}(k+1)^2(k!)^2}{(2k+1+2)!}$$ This can also be written as: $$\frac{4(k+1)^2}{(2k+3)(2k+2)(2k+1)} \cdot \frac{2^{2k}(k!)^2}{(2k+1)!}$$ Now since we assumed the result is true for \(n=k\), $$\frac{4(k+1)^2}{(2k+3)(2k+2)(2k+1)} \cdot \frac{2^{2k}(k!)^2}{(2k+1)!} =\frac{2(k+1)}{2(k+1)+1} \cdot \int_{0}^{1}\left(1-x^{2}\right)^{k} d x$$ If the formula holds, the ratio should be equal. Therefore, we have: $$\frac{2^{2k+2}(k+1)^2(k!)^2}{(2k+1+2)!} =\frac{4(k+1)^2}{(2k+3)(2k+2)(2k+1)} \cdot \frac{2^{2k}(k!)^2}{(2k+1)!}$$ The equation simplifies to: $$\frac{2^{2k+2}(k+1)^2(k!)^2}{(2k+3)(2k+2)(2k+1)(2k+1)!} =\frac{4(k+1)^2}{(2k+3)(2k+2)(2k+1)} \cdot \frac{2^{2k}(k!)^2}{(2k+1)!}$$ We can see that both sides are equal, so our induction is complete. So, we have successfully shown that $$\int_{0}^{1}\left(1-x^{2}\right)^{n} dx = \frac{2^{2n}(n!)^{2}}{(2n+1)!}$$