Definite Integration — Maths STD 12 Science — Question

(d) : Let $I=\int_0^a \sqrt{\frac{a-x}{x}} d x$
Put $x=a \sin ^2 \theta \Rightarrow d x=2 a \sin \theta \cos \theta d \theta$
When $x=0, \theta=0$
When $x=a, \theta=\pi / 2$
$
\begin{aligned}
& \therefore I=\int_0^{\pi / 2} \sqrt{\frac{a\left(1-\sin ^2 \theta\right)}{a \sin ^2 \theta}} \cdot 2 a \sin \theta \cos \theta d \theta \\
& =2 a \int_0^{\pi / 2} \sqrt{\frac{\cos ^2 \theta}{\sin ^2 \theta}} \cdot \sin \theta \cdot \cos \theta d \theta=2 a \int_0^{\pi / 2} \cos ^2 \theta d \theta \\
& =2 a \int_0^{\pi / 2}\left(\frac{1+\cos 2 \theta}{2}\right) d \theta \\
& =2 a\left[\frac{\theta}{2}+\frac{\sin 2 \theta}{4}\right]_0^{\pi / 2}=2 a\left[\frac{\pi}{4}-0+0-0\right]=\frac{a \pi}{2}
\end{aligned}
$
Now, $\frac{a \pi}{2}=\frac{K}{2} \Rightarrow K=a \pi$