$\int_{0}^{a} \sqrt{\frac{x}{a-x}} d x=$

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Aicharya · Accepted Answer

We have, $I=\int_0^a \sqrt{\frac{x}{a-x}} d x$ Put, $x=a \sin ^2 	heta \Rightarrow d x=2 a \sin 	heta \cos 	heta d 	heta$ When, $x=Q, 	heta=0$ and $x=a, 	heta=\pi / 2$ $\begin{aligned} & 	herefore \quad I=\int_0^{\pi / 2} \sqrt{\frac{a \sin ^2 	heta}{a\left(1-\sin ^2 	hetaight)}} 2 a \sin 	heta \cos 	heta d 	heta \ & I=2 a \int_0^{\pi / 2} \frac{\sin 	heta}{\cos 	heta}(\sin 	heta \cos 	heta) d 	heta \ & I=a \int_0^{\pi / 2} 2 \sin ^2 	heta d 	heta \ & I=a \int_0^{\pi / 2}(1-\cos 2 	heta) d 	heta \ & I=a\left[	heta-\frac{\sin 2 	heta}{2}ight]_0^{\pi / 2} \ & I=a\left[\frac{\pi}{2}-0ight]=\left(\frac{\pi}{2}ight) a \end{aligned}$

$\int_{0}^{a} \sqrt{\frac{x}{a-x}} d x=$

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