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$\int \frac{d x}{\sqrt{9 x-4 x^2}}$ equals
(a) $\frac{1}{9} \sin ^{-1}\left(\frac{9 x-8}{8}\right)+C$
(b) $\frac{1}{2} \sin ^{-1}\left(\frac{8 x-9}{9}\right)+C$
(c) $\frac{1}{3} \sin ^{-1}\left(\frac{9 x-8}{8}\right)+C$
(d) $\sin ^{-1}\left(\frac{9 x-8}{9}\right)+C$
$\int \frac{d x}{\sqrt{9 x-4 x^2}}$ equals
(a) $\frac{1}{9} \sin ^{-1}\left(\frac{9 x-8}{8}\right)+C$
(b) $\frac{1}{2} \sin ^{-1}\left(\frac{8 x-9}{9}\right)+C$
(c) $\frac{1}{3} \sin ^{-1}\left(\frac{9 x-8}{8}\right)+C$
(d) $\sin ^{-1}\left(\frac{9 x-8}{9}\right)+C$
Solution:
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Verified Answer
(b)
$\begin{aligned} & \int \frac{\mathrm{dx}}{\sqrt{9 \mathrm{x}-4 \mathrm{x}^2}} \\=& \frac{1}{2} \int \frac{\mathrm{dx}}{\sqrt{\left(\frac{9}{8}\right)^2-\left[x^2-\frac{9}{4} \mathrm{x}+\left(\frac{9}{8}\right)^2\right]}} \\=& \frac{1}{2} \sin ^{-1}\left(\frac{8 x-9}{9}\right)+\mathrm{C} \end{aligned}$
$\begin{aligned} & \int \frac{\mathrm{dx}}{\sqrt{9 \mathrm{x}-4 \mathrm{x}^2}} \\=& \frac{1}{2} \int \frac{\mathrm{dx}}{\sqrt{\left(\frac{9}{8}\right)^2-\left[x^2-\frac{9}{4} \mathrm{x}+\left(\frac{9}{8}\right)^2\right]}} \\=& \frac{1}{2} \sin ^{-1}\left(\frac{8 x-9}{9}\right)+\mathrm{C} \end{aligned}$
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