Evaluate the definite integral: int_{0}^{a} s | vedclass

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(C) To evaluate the integral $I = \int_{0}^{a} \sqrt{\frac{x}{a-x}} \, dx$,substitute $x = a \sin^2 	heta$. Then $dx = 2a \sin 	heta \cos 	heta \,

Vedclass · Accepted Answer

$\frac{\pi}{2} a$

Vedclass · Answer

(C) To evaluate the integral $I = \int_{0}^{a} \sqrt{\frac{x}{a-x}} \, dx$,substitute $x = a \sin^2 	heta$. Then $dx = 2a \sin 	heta \cos 	heta \, d	heta$. When $x = 0$,$	heta = 0$,and when $x = a$,$\sin^2 	heta = 1$,so $	heta = \frac{\pi}{2}$. Substituting these into the integral: $I = \int_{0}^{\pi/2} \sqrt{\frac{a \sin^2 	heta}{a - a \sin^2 	heta}} \cdot (2a \sin 	heta \cos 	heta) \, d	heta$ $I = \int_{0}^{\pi/2} \sqrt{\frac{\sin^2 	heta}{\cos^2 	heta}} \cdot (2a \sin 	heta \cos 	heta) \, d	heta$ $I = \int_{0}^{\pi/2} \frac{\sin 	heta}{\cos 	heta} \cdot (2a \sin 	heta \cos 	heta) \, d	heta$ $I = 2a \int_{0}^{\pi/2} \sin^2 	heta \, d	heta$ Using the identity $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$: $I = 2a \int_{0}^{\pi/2} \frac{1 - \cos 2	heta}{2} \, d	heta = a \int_{0}^{\pi/2} (1 - \cos 2	heta) \, d	heta$ $I = a [	heta - \frac{\sin 2	heta}{2}]_{0}^{\pi/2} = a [(\frac{\pi}{2} - 0) - (0 - 0)] = \frac{\pi}{2} a$. Thus,the correct option is $(C)$.

Evaluate the definite integral: $\int_{0}^{a} \sqrt{\frac{x}{a-x}} \, dx$

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