Trigonometric substitution in the integral $\int x^2 (1-x^2)^{\frac{9}{2}} \ \mathrm dx$ I'm trying to solve
$$\int_{-1}^{1} x^2(1-x^2)^{\frac{9}{2}} \, dx$$
The hint said to use the substitution $x=\sin y$
I got $$\int_{-\pi/2}^{\pi/2} \sin^2y \cos^{\frac{11}{2}} y \,dy$$

As said in comments, the antiderivative reduces to the evaluation of $$I=\int \cos^{10}(x)dx-\int \cos^{12}(x)dx=J_{10}-J_{12}$$ with $$J_n=\int \cos^{n}(x)dx$$ The $J_n$ terms can easily be computed since we can easily establish (performing two integrations by parts) the recurrence relation $$J_n=\frac 1n \cos^{n-1}(x)\sin(x)-\frac{n-1}{n}J_{n-2}$$ with $$J_0=x$$ $$J_1=-\sin(x)$$ Integrating between the given bounds $(-\pi/2,\pi/2)$, most terms disappear because of the cosines and  we can then establish that $$J_n=\frac{\sqrt{\pi } \Gamma \left(\frac{n+1}{2}\right)}{\Gamma
   \left(\frac{n}{2}+1\right)}$$ So $J_{10}=\frac{63 \pi }{256}$ and $J_{12}=\frac{231 \pi }{1024}$ and then the result Adhvaitha gave.
Notice that if $n=2m$ this simplify further $$J_{2m}=\frac{\sqrt{\pi } \Gamma \left(m+\frac{1}{2}\right)}{m!}$$ with $\Gamma \left(\frac{1}{2}\right)=\sqrt{\pi}$ and $\Gamma \left(m+\frac{1}{2}\right)=\left(m-\frac{1}{2}\right) \Gamma
   \left(m-\frac{1}{2}\right)$.
