How to solve $\int ^{1}_{-1}\frac {x^{2n}}{\sqrt {1-x^{2}}}dx?$ I couldn't solve this.
$$\int ^{1}_{-1}\dfrac {x^{2n}}{\sqrt {1-x^{2}}}dx$$
I thought that like the following.
$$\int ^{1}_{-1}\dfrac {x^{2n}}{\sqrt {1-x^{2}}}dx=\int ^{1}_{-1}\dfrac {1-\left( 1-x^{2n}\right) }{\sqrt {1-x^{2}}}dx\\=\int ^{1}_{-1}\dfrac {1-\sqrt {\left( 1-x^{2n}\right) ^{2}}}{\sqrt {1-x^{2}}}dx\\=\int ^{1}_{-1}\dfrac {1}{\sqrt {1-x^{2}}}dx-\int ^{1}_{-1}\dfrac {\sqrt {\left( 1-x^{2n}\right) ^{2}}}{\sqrt {1-x^{2}}}dx$$
I don't know what to do next.  Maybe the procedure so far is wrong. Please tell me how to solve.
 A: Using
$$I_n=\int ^{1}_{-1}\dfrac {x^{2n}}{\sqrt {1-x^{2}}}\,dx=2\int_0^{\frac{\pi}{2}}\sin^{2n}(t)\,dt=\sqrt{\pi }\frac{ \Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1)}=4^{-n} \binom{2 n}{n}\pi$$
A: Either set $x= \sin y$ or  then we have $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^{2n} y =2\int_0^{\frac{\pi}{2}}\sin^{2n}y dy\\=\pi{2n\choose n}\frac{1}{4^n}  $$  notice that the last integral we have is Walli's Integral.
Without subbing $$\int_{-1}^{1}x^{2n} (1-x^2)^{\frac{-1}{2}} dx=2\int_0^1x^{2n} \left(\sum_{k=0}^{\infty} (-1)^k  {-\frac{1}{2}\choose k} x^{2k}\right) dx \\=2\sum_{k=0}^{\infty}(-1)^{k}{-\frac{1}{2}\choose k}\int_0^1x^{2(n+k)}dx=\sum_{k=0}^{\infty}(-1)^k{-\frac{1}{2} \choose k} \frac{2}{2n+2k+1}$$
A: Let $x=\sin\theta\implies dx=\cos\theta \ d\theta$
$$\int ^{1}_{-1}\dfrac {x^{2n}}{\sqrt {1-x^{2}}}dx=2\int ^{1}_{0}\dfrac {x^{2n}}{\sqrt {1-x^{2}}}dx$$
$$=2\int ^{\pi/2}_{0}\dfrac {\sin^{2n}\theta}{\cos\theta}\cos\theta \ d\theta$$
$$=2\int ^{\pi/2}_{0}\sin^{2n}\theta\ d\theta$$
Using formula $\color{blue}{\int_0^{\pi/2}\sin^m\theta\cos^n\theta\ d\theta=\dfrac{\Gamma(\frac{m+1}{2})\Gamma(\frac{n+1}{2})}{2\Gamma(\frac{m+n+2}{2})}} $,
$$=2\frac{\Gamma(\frac{2n+1}{2})\Gamma(\frac{0+1}{2})}{2\Gamma(\frac{2n+0+2}{2})}$$
$$=\frac{\Gamma(n+\frac{1}{2})\sqrt{\pi}}{\Gamma(n+1)}$$
