int frac{dx}{sqrt{9x-4x^{2}}} equals | vedclass

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Question

(D) To evaluate the integral $I = \int \frac{dx}{\sqrt{9x-4x^{2}}}$,we complete the square inside the square root.$I = \int \frac{dx}{\sqrt{-4(x^{2}-\

Vedclass · Accepted Answer

$\frac{1}{2} \sin^{-1}\left(\frac{8x-9}{9}\right)+C$

Vedclass · Answer

(D) To evaluate the integral $I = \int \frac{dx}{\sqrt{9x-4x^{2}}}$,we complete the square inside the square root.$I = \int \frac{dx}{\sqrt{-4(x^{2}-\frac{9}{4}x)}}$$I = \int \frac{dx}{\sqrt{-4(x^{2}-\frac{9}{4}x + \frac{81}{64} - \frac{81}{64})}}$$I = \int \frac{dx}{\sqrt{-4[(x-\frac{9}{8})^{2} - (\frac{9}{8})^{2}]}}$$I = \int \frac{dx}{2\sqrt{(\frac{9}{8})^{2} - (x-\frac{9}{8})^{2}}}$Using the standard integral formula $\int \frac{dy}{\sqrt{a^{2}-y^{2}}} = \sin^{-1}(\frac{y}{a}) + C$,where $a = \frac{9}{8}$ and $y = x - \frac{9}{8}$:$I = \frac{1}{2} \sin^{-1}\left(\frac{x-\frac{9}{8}}{\frac{9}{8}}\right) + C$$I = \frac{1}{2} \sin^{-1}\left(\frac{8x-9}{9}\right) + C$Thus,the correct option is $D$.

$\int \frac{dx}{\sqrt{9x-4x^{2}}}$ equals

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