Integrate:$ \int_{0}^{1} x^{1/2}(1-x^{2})^{3/2}dx$ Integrate:
$\int_{0}^{1} x^{1/2}(1-x^{2})^{3/2}dx$
I tried substituting $x=\sin \theta$ and ended up with $\int_{0}^{\pi/2} \sqrt{\sin\theta}\cos ^{4}\theta d\theta$. Further substitution made it more complex.
After this what should I do?
 A: See the identities here: https://en.wikipedia.org/wiki/Beta_function
The key identities related to your problem are
\begin{align}
B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)} &= \int_0^1 t^{x-1}(1-t)^{y-1} \;dt \\
&= n \int_0^1 t^{nx-1}(1-t^n)^{y-1} \;dt \\
&= 2 \int_0^{\pi/2} (\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\;d\theta.
\end{align}
So your integral is
$$ \frac12 \cdot 2 \int_0^1 x^{2\left(\frac34\right)-1} (1 - x^2)^{\left(\frac52\right)-1} \;dx = \frac12 B\left(\frac34,\frac52\right) = \frac{\Gamma\left(\frac34\right) \Gamma\left(\frac52\right)}{2\Gamma\left(\frac{13}4\right)} = \frac{3\sqrt\pi}8 \frac{\Gamma\left(\frac34\right)}{\Gamma\left(\frac{13}4\right)}.$$
A: $$I=\int_0^1 x^{1/2}(1-x^2)^{3/2}dx$$
$u=x^2\Rightarrow dx=\frac{du}{2x}$ now sub in:
$$I=\int\limits_{u=0}^1u^{1/4}(1-u)^{3/2}\frac{du}{2u^{1/2}}=\frac12\int_0^1u^{-1/4}(1-u)^{3/2}du$$
Now the definition of the beta function is:
$$B(x,y)=\int_0^1t^{x-1}(1-t)^{y-1}dt\tag{1}$$
$$B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\tag{2}$$
where $\Gamma$ is the Gamma function, an analytic continuation of the factorial function.
