Enfoque de la función beta
Sustituyendo$x\mapsto2x+3$, $$ \begin{align}
&\int_3^5\frac{x^2\,\mathrm{d}x}{\sqrt{(x-3)(5-x)}}\\
&=\int_0^1\frac{(2x+3)^2\,\mathrm{d}x}{\sqrt{x(1-x)}}\\
&=\int_0^1\frac{(3(1-x)+5x)^2\,\mathrm{d}x}{\sqrt{x(1-x)}}\\
&=9\int_0^1\frac{(1-x)^2\,\mathrm{d}x}{\sqrt{x(1-x)}}
+30\int_0^1\frac{(1-x)x\,\mathrm{d}x}{\sqrt{x(1-x)}}
+25\int_0^1\frac{x^2\,\mathrm{d}x}{\sqrt{x(1-x)}}\\
&=9\operatorname{B}\left(\frac52,\frac12\right)
+30\operatorname{B}\left(\frac32,\frac32\right)
+25\operatorname{B}\left(\frac12,\frac52\right)\\
&=9\cdot\frac38\pi
+30\cdot\frac18\pi
+25\cdot\frac38\pi\\
&=\frac{33}2\pi
\end {align} $$ usando la función Beta .
Subsidio trigonométrico
Después de la sustitución$x\mapsto2x+3$, podemos usar$x\mapsto\sin^2(x)$ $$ \begin{align}
&\int_3^5\frac{x^2\,\mathrm{d}x}{\sqrt{(x-3)(5-x)}}\\
&=9\int_0^1\frac{(1-x)^2\,\mathrm{d}x}{\sqrt{x(1-x)}}
+30\int_0^1\frac{(1-x)x\,\mathrm{d}x}{\sqrt{x(1-x)}}
+25\int_0^1\frac{x^2\,\mathrm{d}x}{\sqrt{x(1-x)}}\\
&=9\int_0^{\pi/2}2\cos^4(x)\,\mathrm{d}x
+30\int_0^{\pi/2}2\sin^2(x)\cos^2(x)\,\mathrm{d}x
+25\int_0^{\pi/2}2\sin^4(x)\,\mathrm{d}x\\
&=9\cdot\frac38\pi
+30\cdot\frac18\pi
+25\cdot\frac38\pi\\
&=\frac{33}2\pi
\end {align} $$