int frac{1}{x^2 sqrt{1-x^2}} cdot d x = dots | vedclass

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Question

(A) To solve the integral $I = \int \frac{1}{x^2 \sqrt{1-x^2}} \cdot dx$,we use the substitution $x = \sin 	heta$. Then,$dx = \cos 	heta \cdot d	he

Vedclass · Accepted Answer

$-\frac{\sqrt{1-x^2}}{x}$

Vedclass · Answer

(A) To solve the integral $I = \int \frac{1}{x^2 \sqrt{1-x^2}} \cdot dx$,we use the substitution $x = \sin 	heta$. Then,$dx = \cos 	heta \cdot d	heta$. Substituting these into the integral: $I = \int \frac{\cos 	heta \cdot d	heta}{\sin^2 	heta \sqrt{1-\sin^2 	heta}} = \int \frac{\cos 	heta \cdot d	heta}{\sin^2 	heta \cdot \cos 	heta} = \int \frac{1}{\sin^2 	heta} \cdot d	heta = \int \csc^2 	heta \cdot d	heta$. The integral of $\csc^2 	heta$ is $-\cot 	heta + C$. Since $x = \sin 	heta$,we have $\sin 	heta = x$. Then $\cos 	heta = \sqrt{1-x^2}$,so $\cot 	heta = \frac{\cos 	heta}{\sin 	heta} = \frac{\sqrt{1-x^2}}{x}$. Thus,$I = -\frac{\sqrt{1-x^2}}{x} + C$. Therefore,the correct option is $A$.

$\int \frac{1}{x^2 \sqrt{1-x^2}} \cdot d x = \dots + C$. Where,$(0 < |x| < 1)$.

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