\(\int_0^a \sqrt{\frac{a-x}{x}} d x=\frac{k}{2}\), then \(k=\)

We have \(\int_0^3 \sqrt{\frac{a-x}{x}} d x=\frac{k}{2}\)
Put \(x=a \sin ^2 \theta \Rightarrow d x=a(2 \sin \theta) \cos \theta d \theta~=\) \(2 a \sin \theta \cos \theta d \theta\)
When \(\mathrm{x}=0, \theta=0\) and when \(\mathrm{x}=\mathrm{a}, \theta=\frac{\pi}{2}\)
\( \therefore \int_0^{\frac{\pi}{2}} \sqrt{\frac{\mathrm{a}-\mathrm{a} \sin ^2 \theta}{\mathrm{a} \sin ^2 \theta}}(2 \mathrm{a} \sin \theta \cos \theta) \mathrm{d} \theta=\frac{\mathrm{k}}{2} \)
\( \int_0^{\frac{\pi}{2}} \sqrt{\frac{\cos ^2 \theta}{\sin ^2 \theta}} 2(\mathrm{a} \sin \theta \cos \theta) \mathrm{d} \theta=\frac{\mathrm{k}}{2} \Rightarrow \int_0^{\frac{\pi}{2}}\) \(\sqrt{\frac{\cos \theta}{\sin \theta}} 2(\mathrm{a} \sin \theta \cos \theta) \mathrm{d} \theta=\frac{\mathrm{k}}{2} \)
\( \int_0^{\frac{\pi}{2}} 2 \mathrm{a} \cos \theta^2 \mathrm{~d} \theta=\frac{\mathrm{k}}{2} \Rightarrow 2 \mathrm{a} \int_0^{\frac{\pi}{2}} \frac{1+\cos 2 \theta}{2} \mathrm{~d} \theta=\frac{\mathrm{k}}{2} \)
\( \int_0^{\frac{\pi}{2}} 2 \mathrm{a} \cos \theta^2 \mathrm{~d} \theta=\frac{\mathrm{k}}{2} \Rightarrow 2 \mathrm{a} \int_0^{\frac{\pi}{2}} \frac{1+\cos 2 \theta}{2} \mathrm{~d} \theta=\frac{\mathrm{k}}{2} \)
\( \mathrm{a}\left[\left(\frac{\pi}{2}+0\right)-(0+0)\right]=\frac{\mathrm{k}}{2} \Rightarrow \mathrm{a} \frac{\pi}{2}=\frac{\mathrm{k}}{2} \Rightarrow \mathrm{k}=\pi \mathrm{a}\)