Evaluate the given definite integral How to evaluate the integral $$\int_0^1 (1-x^{2/3})^{3/2} \, dx.$$ Does some substitution work?
 A: $$
\begin{align}
\int_0^1(1-x^{2/3})^{3/2}\,\mathrm{d}x
&=\int_0^1(1-u)^{3/2}\,\mathrm{d}u^{3/2}\\
&=\frac32\int_0^1(1-u)^{3/2}u^{1/2}\,\mathrm{d}u\\
&=\frac32\mathrm{B}\left(\frac52,\frac32\right)\\
&=\frac32\frac{\Gamma\left(\frac52\right)\Gamma\left(\frac32\right)}{\Gamma(4)}\\
&=\frac32\frac{\frac32\frac12\Gamma\left(\frac12\right)\frac12\Gamma\left(\frac12\right)}{\Gamma(4)}\\
&=\frac{3\pi}{32}
\end{align}
$$

Alternatively, we can use
$$
\begin{align}
\int_0^{\pi/2}\cos^{2n}(x)\,\mathrm{d}x
&=\frac14\int_0^{2\pi}\cos^{2n}(x)\,\mathrm{d}x\\
&=\frac14\int_0^{2\pi}\left(\frac{e^{ix}+e^{-ix}}{2}\right)^{2n}\,\mathrm{d}x\\
&=\frac14\int_0^{2\pi}4^{-n}\sum_{k=0}^{2n}\binom{2n}{k}e^{i(2n-2k)x}\,\mathrm{d}x\\
&=\frac\pi2\binom{2n}{n}4^{-n}
\end{align}
$$
and
$$
\begin{align}
\int_0^1(1-x^{2/3})^{3/2}\,\mathrm{d}x
&=\int_0^1(1-u)^{3/2}\,\mathrm{d}u^{3/2}\\
&=\int_0^1(1-v^2)^{3/2}\,\mathrm{d}v^3\\
&=3\int_0^1(1-v^2)^{3/2}v^2\,\mathrm{d}v\\
&=3\int_0^{\pi/2}\cos^4(\theta)\sin^2(\theta)\,\mathrm{d}\theta\\
&=3\int_0^{\pi/2}\left[\cos^4(\theta)-\cos^6(\theta)\right]\,\mathrm{d}\theta\\
&=\frac{3\pi}{2}\left[\binom{4}{2}4^{-2}-\binom{6}{3}4^{-3}\right]\\
&=\frac{3\pi}{32}
\end{align}
$$
A: Hint: substitute $u=x^{1/3}$. This puts the integral in a form amenable to a simple trigonometric substitution.
Substituting $u=x^{1/3}$,
$$I=\int_{0}^{1}(1-x^{2/3})^{3/2}dx=3\int_{0}^{1}u^2\,(1-u^2)^{3/2}\,du.$$
Substituting $u=\sin{\theta}$,
$$I=3\int_{0}^{1}u^2\,(1-u^2)^{3/2}\,du=3\int_{0}^{\pi/2}\sin^2{\theta}\,(1-\sin^2{\theta})^{3/2}\cos{\theta}\,d\theta\\
=3\int_{0}^{\pi/2}\sin^2{\theta}\,\cos^4{\theta}\,d\theta\\
=3\int_{0}^{\pi/2}\cos^4{\theta}\,d\theta-3\int_{0}^{\pi/2}\cos^6{\theta}\,d\theta\\
=\frac{9\pi}{16}-\frac{15\pi}{32}\\
=\frac{3\pi}{32}.$$
A: $$\int_0^1\Big(1-\sqrt[n]x\Big)^m~=~\int_0^1\Big(1-\sqrt[m]x\Big)^n~=~\frac{m!\cdot n!}{(m+n)!}~=~{m+n\choose n}^{-1}~=~{m+n\choose m}^{-1}$$

$\qquad\quad$ In this case, $m=n=\dfrac32$ , with $\Gamma\bigg(\dfrac12\bigg)=\sqrt\pi,~$ and $n!=\Gamma(n+1)=n\cdot\Gamma(n)$.
A: You can also  use $x=\sin^3\theta$ and $dx=3\sin^2\theta\cos\theta\ d\theta$, then
$$
\begin{align}
\int_0^1\left(1-x^{\large\frac23}\right)^{\large\frac32}\ dx&=\int_0^{\large\frac\pi2}\left(1-\sin^2\theta\right)^{\large\frac32}\cdot 3\sin^2\theta\ \cos\theta\ d\theta\\
&=3\int_0^{\large\frac\pi2}\cos^4\theta(1-\cos^2\theta)\ d\theta\\
&=3\int_0^{\large\frac\pi2}(\cos^4\theta-\cos^6\theta)\ d\theta.
\end{align}
$$
The last integral can be solved by using integration by reduction formula, we will obtain
$$
\begin{align}
\int_0^1\left(1-x^{\large\frac23}\right)^{\large\frac32}\ dx&=3\int_0^{\large\frac\pi2}\cos^4\theta\ d\theta-3\int_0^{\large\frac\pi2}\cos^6\theta\ d\theta\\
&=3\int_0^{\large\frac\pi2}\cos^4\theta\ d\theta-3\left(\left.\frac16\cos^5\theta\sin\theta\right|_0^{\large\frac\pi2}+\frac56\int_0^{\large\frac\pi2}\cos^4\theta\ d\theta\right)\\
&=\frac12\int_0^{\large\frac\pi2}\cos^4\theta\ d\theta.
\end{align}
$$
The rest I leave it to you.
