Another way to calculate the integral $ \int_{0}^{1} (x-x^{2})^{3/2} dx $

Question

$$\int_{0}^{1} (x-x^{2})^{3/2} dx=\dfrac{1}{8}\int_{0}^{1}[1-(2x-1)^{2}]^{3/2}dx=\dfrac{1}{8}\int_{-\pi/2}^{\pi/2}\dfrac{cos^{4}\theta}{2}d\theta=\dfrac{1}{8}\int_{0}^{\pi/2}cos^{4}d\theta=\dfrac{1}{8}(\dfrac{3\pi}{16})=\dfrac{3\pi}{128}$$ The change of variable $2x-1 = sin\theta$ was made.

How could it be integrated without the "trick" $(x-x^{2})^{3/2}=\dfrac{1}{8}[1-(2x-1)^{2}]^{3/2}$?

Answer 1

There is also the following way.

Let $x=\cos^2t$, where $t\in\left[0,\frac{\pi}{2}\right]$.

Thus, $$\int\limits_0^1(x-x^2)^{\frac{3}{2}}dx=\int_{\frac{\pi}{2}}^0\sin^3t\cos^3t(-2\sin{t}\cos{t})dt=2\int\limits_0^{\frac{\pi}{2}}\sin^4t\cos^4tdt=$$ $$=\frac{1}{8}\int\limits_0^{\frac{\pi}{2}}\sin^42tdt=\frac{1}{32}\int\limits_0^{\frac{\pi}{2}}(1-\cos4t)^2dt=\frac{1}{32}\int\limits_0^{\frac{\pi}{2}}\left(1-2\cos4t+\frac{1+\cos8t}{2}\right)dt=$$ $$=\frac{1}{32}\left(1+\frac{1}{2}\right)\frac{\pi}{2}=\frac{3\pi}{128}.$$

Answer 2

See this, $$\int x^{3/2} (1-x)^{3/2} dx$$ $$=B(5/2,5/2)=\frac{3 \pi}{128}$$

$B(x,y)$ is Beta function.