Solve $ \int_0^a x^4 \sqrt{a^2-x^2}dx$ Solve $$ \int_0^a x^4 \sqrt{a^2-x^2}dx.$$


Substitution: $x = a\sin t \Rightarrow dx = a\cos t dt$

$$ \int_0^a x^4 \sqrt{a^2-x^2}dx = a^6 \int_0^{\pi/2}\sin^4t\cos^2 tdt = \frac{a^6}6 \frac{\Gamma(5/2)\Gamma(3/2)}{\Gamma(3)} = \frac{a^6 \pi}{288}.$$

When evaluating this integral in Wolfram Alpha, I get another result. Also plugging in values and checking if they correspond to the answer I would get with my result, give contradictions. Where is my mistake?

Thanks.



EDIT: I found my mistake. Just had  a little brain fart: $\Gamma(3) = 2!$ and not $3!$. Thus we get $$\frac{a^6}{12}\cdot \frac{3\pi}{8} = \frac{a^6 \pi}{32},$$ which is correct. Thank you for your time and responses!
A direct path to the Beta function can be made by considering the substitution $x = a \, \sqrt{u}$ in the integral:
\begin{align}
I &amp;= \int_{0}^{a} x^{4} \, (a^2 - x^2)^{1/2} \, dx \\
&amp;= \int_{0}^{1} a^4 \, u^2 \, (a^2 - a^2 u)^{1/2} \, \frac{a \, du}{2 \, \sqrt{u}} \\
&amp;= \frac{a^6}{2} \, \int_{0}^{1} u^{3/2} \, (1-u)^{1/2} \, du \\
&amp;= \frac{a^6}{2} \, B\left(\frac{5}{2}, \frac{3}{2}\right) \\
&amp;= \frac{a^6 \, \pi}{32}.
\end{align}
Hint: Substitute $$\sqrt{a^2-x^2}=xt\pm a$$ the so-called Euler Substitution.