Integral of $\sin(t)(1-\sin(t))^{3/2}$ I am trying to calculate
$$\int\limits_{-\pi/2}^{\pi/2} \sin(t)(1-\sin(t))^{3/2} \; dt$$
Wolfram Alpha suggests a double substitution to get a polynomial, integrating, then resubstituting to get a very complex term. Is there a more elegant way?

Substitute $\sin(t) = x$
$$\implies dx = \cos(t) dt$$
$$\iff dt = \frac{dx}{\cos(t)} = \frac{dx}{\sqrt{1-\sin(t)^2}} = \frac{dx}{\sqrt{1-x^2}}$$.
$$\int_{-\pi/2}^{\pi/2} \sin(t)\cdot(1-\sin(t))^{3/2} dt$$
$$=\int_{-1}^1 \frac{x(1-x)^{3/2}}{\sqrt{1-x^2}} dx$$
$$=\int_{-1}^1 \frac{x(1-x)^{3/2}}{\sqrt{1+x}\sqrt{1-x}} dx$$
$$=\int_{-1}^1 \frac{x(1-x)}{\sqrt{1+x}} dx$$
substitute $1+x=y$
$$=\int_0^2 \frac{(y-1)(2-y)}{\sqrt{y}} dy$$
expand powers to yield elementary integrals
$$= -\frac{8\sqrt{2}}{5}$$
