How to Integrate this function $\int(1-x^2)^ndx$

The actual question was to find the area bounded by the curve $y=\int_{-1}^{1}(1-x^2)^ndx$ and coordinate axes. But I haven't came across these type of problems with power $n$.

Solution 1:

By parts,

$$I_n=\int(1-x^2)^ndx=x(1-x^2)^n+2n\int x^2(1-x^2)^{n-1}dx.$$

But $x^2=1-(1-x^2)$, so that

$$I_n=x(1-x^2)^n+2nI_{n-1}-2nI_n,$$ and

$$I_n=\frac{x(1-x^2)^n}{2n+1}+\frac{2n}{2n+1}I_{n-1}.$$

With $I_0=x$, you can compute for any $n$.

When evaluating the definite integral, the first term vanishes and

$$I_n=\frac{2(2n)!!}{(2n+1)!!}.$$

Solution 2:

Using the binomial theorem, $(1-x^2)^n=\sum_{k=0}^n\binom{n}{k}(-1)^kx^{2k}$ and therefore

$$\int (1-x^2)^n\,dx=\sum_{k=0}^n\binom{n}{k}(-1)^k\frac{x^{2k+1}}{2k+1}+C$$

and

$$\int_{-1}^1 (1-x^2)^n\,dx=\sum_{k=0}^n\binom{n}{k}(-1)^k\frac{2}{2k+1}$$

ALTERNATIVE APPROACH TO EVALUATE THE DEFINITE INTEGRAL:

First, note that $\int_{-1}^1 (1-x^2)^n\,dx=2\int_0^1 (1-x^2)^n\,dx$. Next, enforce the substitution $x\to \sqrt{x}$ to obtain

$$\begin{align} \int_{-1}^1 (1-x^2)^n\,dx&=2\int_0^1 (1-x)^n x^{-1/2}\,dx\\\\ &=2\,B(n+1,1/2)\\\\ &=2\frac{\Gamma(n+1)\Gamma(1/2)}{\Gamma(n+3/2)}\\\\ &=2\frac{n!\Gamma(1/2)}{(n+1/2)!\Gamma(1/2)}\\\\ &=2\frac{n!}{\frac32\,\frac52\,\frac72\,\cdots\,\frac{2n+1}{2}}\\\\ &=2\frac{2^n\,n!}{(2n+1)!!}\\\\ &=2\frac{(2n)!!}{(2n+1)!!} \end{align}$$

as expected!