Let,

$ I_1=\int_{0}^{3}x^{3/2}(3-x)^{-1/2}dx$

put,

$x=3t \hspace{0.3cm}=\gt\hspace{0.3cm} dx=3dt$

when,

$x=0,t=0,x=3,t=1$

$\int_{0}^{1}{(3t)^{3/2}}(3-3t)^{-1/2}3dt = 3^{3/2-1/2+1}\int_{0}^{i}t^{3/2}(1-t)^{-1/2}dt$

$\hspace{4.4cm}= 9\beta(\frac{5}{2},\frac{1}{2})=9\frac{\sqrt{5/2}\sqrt{1/2}}{\sqrt{\frac{5}{2}+\frac{1}{2}}}$

$\hspace{4.4cm}= 9*\frac{(3/2) \,(1/2)\sqrt{1/2}\sqrt{1/2}}{\sqrt{3}}$

$\hspace{4.4cm}= \frac{27}{4} \frac{\sqrt{\pi}}{2}\,\sqrt{\pi}$

$\hspace{4.4cm}= \frac{27\pi}{8} $

Let,

$I_2=\int_{0}^{1}\frac{dx}{\sqrt{1-x^{1/4}}} $

Put,

$x^{1/4}=t \hspace{0.3cm}=\gt \hspace{0.3cm}x = t^4 \hspace{0.3cm}=\gt \hspace{0.3cm} dx=4t^3dt $

$\int_{0}^{1}\frac{4t^3dt}{\sqrt{1-t}} = 4\int_{0}^{1} t^3(1-t)^{-1/2}dt $

$\hspace{1.5cm}= 4\beta(4,1/2)= \frac{4\sqrt{4}\sqrt{1/2}}{\sqrt{4+\frac{1}{2}}}$

$\hspace{1.5cm}= \frac{4(3!)\sqrt{\pi}}{\sqrt{9/2}} = \frac{24\sqrt{\pi}}{\frac{7}{2} *\frac{5}{2} *\frac{3}{2}*\frac{1}{2}\sqrt{\pi}}$

$\hspace{1.5cm}= \frac{24\sqrt{\pi}}{105/16 \sqrt{\pi}} =\frac{128}{35} $

$I_1*I_2=\frac{27}{8}\pi=\frac{128}{35}=\frac{432}{35}\pi$