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 $$\int_{-\pi/2}^{\pi/2} \sin^2y \cos^{\frac{11}{2}} y \,dy$$
 A: Setting $x=\sin(t)$, we obtain
$$I = \int_{-\pi/2}^{\pi/2} \sin^2(t) \cos^9(t) \cos(t) dt = \int_{-\pi/2}^{\pi/2} \sin^2(t) \cos^{10}(t) dt = \int_{-\pi/2}^{\pi/2} \cos^{10}(t)dt - \int_{-\pi/2}^{\pi/2}\cos^{12}(t)dt$$
From here, we have
$$\int_{-\pi/2}^{\pi/2} \cos^{2m}(t)dt = \dfrac{\pi}{2^{2m}}\dbinom{2m}m$$
Hence,
$$I = \dfrac{\pi}{2^{10}}\dbinom{10}5 - \dfrac{\pi}{2^{12}}\dbinom{12}6 = \dfrac{21\pi}{1024}$$
A: 
$$x=\sin y$$


$$\int x^2 ( 1-x^2)^{\frac{9}{2}}\,dx = \int x^2 (\sqrt{1-x^2})^9 \,dx= \int \sin^2 y (\sqrt{1-sin^2 y})^9 \cos y\, dy $$
$$= \int \sin^2 y (\sqrt\cos^2 y)^9 \cos y\, dy= \int \sin^2 y (\cos y)^9 \cos y\, dy$$
$$\int \sin^2 y \cos^{10} y \,dy$$
A: 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 $\Gamma \left(m+\frac{1}{2}\right)=\left(m-\frac{1}{2}\right) \Gamma
   \left(m-\frac{1}{2}\right)$.
