91Ó°ÊÓ

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 \(v = x\)

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: $$\int_{0}^{1}(1-x^2)^ndx= 2n \int_{0}^{1}x^{2n} dx$$

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

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 $$\int_{0}^{1}\left(1-x^{2}\right)^{n} dx = \frac{2^{2n}(n!)^{2}}{(2n+1)!}$$