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 $$\int_{-\pi/2}^{\pi/2} \sin^2y \cos^{\frac{11}{2}} y \,dy$$


$$x=\sin y$$


$$\int x^2 ( 1-x^2)^{\frac{9}{2}}\,dx = \int x^2 (\sqrt{1-x^2})^9 \,dx= \int \sin^2 y (\sqrt{1-sin^2 y})^9 \cos y\, dy $$
$$= \int \sin^2 y (\sqrt\cos^2 y)^9 \cos y\, dy= \int \sin^2 y (\cos y)^9 \cos y\, dy$$
$$\int \sin^2 y \cos^{10} y \,dy$$
