The value of the definite integral intlimits_ | vedclass

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Question

(C) Let $I = \int\limits_0^{\frac{1}{{\sqrt 2 }}} {\frac{{{x^2}dx}}{{\sqrt {1 - {x^2}} \,(1 + \sqrt {1 - {x^2}} )}}} $. Substitute $x = \sin 	heta$,t

Vedclass · Accepted Answer

$\frac{\pi }{4} - \frac{1}{{\sqrt 2 }}$

Vedclass · Answer

(C) Let $I = \int\limits_0^{\frac{1}{{\sqrt 2 }}} {\frac{{{x^2}dx}}{{\sqrt {1 - {x^2}} \,(1 + \sqrt {1 - {x^2}} )}}} $. Substitute $x = \sin 	heta$,then $dx = \cos 	heta \, d	heta$. When $x = 0$,$	heta = 0$. When $x = \frac{1}{\sqrt{2}}$,$	heta = \frac{\pi}{4}$. Substituting these into the integral: $I = \int\limits_0^{\frac{\pi}{4}} \frac{\sin^2 	heta \cos 	heta \, d	heta}{\cos 	heta (1 + \cos 	heta)} = \int\limits_0^{\frac{\pi}{4}} \frac{\sin^2 	heta}{1 + \cos 	heta} \, d	heta$. Using the identity $\sin^2 	heta = 1 - \cos^2 	heta = (1 - \cos 	heta)(1 + \cos 	heta)$: $I = \int\limits_0^{\frac{\pi}{4}} \frac{(1 - \cos 	heta)(1 + \cos 	heta)}{1 + \cos 	heta} \, d	heta = \int\limits_0^{\frac{\pi}{4}} (1 - \cos 	heta) \, d	heta$. Integrating,we get: $I = [	heta - \sin 	heta]_0^{\frac{\pi}{4}} = (\frac{\pi}{4} - \sin \frac{\pi}{4}) - (0 - \sin 0) = \frac{\pi}{4} - \frac{1}{\sqrt{2}}$. Thus,the correct option is $C$.

The value of the definite integral $\int\limits_0^{\frac{1}{{\sqrt 2 }}} {\frac{{{x^2}dx}}{{\sqrt {1 - {x^2}} \,(1 + \sqrt {1 - {x^2}} )}}} $ is

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