int frac{dx}{sqrt{x-x^2}} is equal to | vedclass

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(A) Let $I = \int \frac{dx}{\sqrt{x-x^2}}$.We can rewrite the denominator as $\sqrt{x(1-x)} = \sqrt{x} \sqrt{1-x}$.So,$I = \int \frac{dx}{\sqrt{x} \sq

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$2 \sin^{-1} \sqrt{x} + C$

Vedclass · Answer

(A) Let $I = \int \frac{dx}{\sqrt{x-x^2}}$.We can rewrite the denominator as $\sqrt{x(1-x)} = \sqrt{x} \sqrt{1-x}$.So,$I = \int \frac{dx}{\sqrt{x} \sqrt{1-x}}$.Let $\sqrt{x} = \sin 	heta$. Then $x = \sin^2 	heta$,which implies $dx = 2 \sin 	heta \cos 	heta \, d	heta$.Substituting these into the integral:$I = \int \frac{2 \sin 	heta \cos 	heta \, d	heta}{\sin 	heta \sqrt{1-\sin^2 	heta}}$$I = \int \frac{2 \sin 	heta \cos 	heta \, d	heta}{\sin 	heta \cos 	heta}$$I = \int 2 \, d	heta = 2	heta + C$.Since $\sin 	heta = \sqrt{x}$,we have $	heta = \sin^{-1} \sqrt{x}$.Therefore,$I = 2 \sin^{-1} \sqrt{x} + C$.

$\int \frac{dx}{\sqrt{x-x^2}}$ is equal to

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