$\int \frac{1}{\sqrt{9 x-4 x^{2}}} d x$ के बराबर है:
Exercise-7.4-25
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$\int \frac{1}{\sqrt{9 x-4 x^{2}}} d x$ $=\frac{1}{\sqrt{4}} \int \frac{1}{\sqrt{\frac{9}{4} x-x^{2}}} d x$
$=\frac{1}{2} \int \frac{1}{\sqrt{-\left[x^{2}-\frac{9}{4} x+\left(\frac{9}{8}\right)^{2}\right]+\left(\frac{9}{8}\right)^{2}}} d x$ $=\frac{1}{2} \int \frac{1}{\sqrt{\left(\frac{9}{8}\right)^{2}-\left(x-\frac{9}{8}\right)^{2}}} d x$
$=\frac{1}{2} \sin ^{-1}\left(\frac{x-\frac{9}{8}}{\frac{9}{8}}\right)$ + C [$\because$ $\int \frac{d x}{\sqrt{a^{2}-x^{2}}}$ $=\sin ^{-1}\left(\frac{x}{a}\right)$]
$=\frac{1}{2} \sin ^{-1}\left(\frac{8 x-9}{9}\right)$ + C
$=\frac{1}{2} \int \frac{1}{\sqrt{-\left[x^{2}-\frac{9}{4} x+\left(\frac{9}{8}\right)^{2}\right]+\left(\frac{9}{8}\right)^{2}}} d x$ $=\frac{1}{2} \int \frac{1}{\sqrt{\left(\frac{9}{8}\right)^{2}-\left(x-\frac{9}{8}\right)^{2}}} d x$
$=\frac{1}{2} \sin ^{-1}\left(\frac{x-\frac{9}{8}}{\frac{9}{8}}\right)$ + C [$\because$ $\int \frac{d x}{\sqrt{a^{2}-x^{2}}}$ $=\sin ^{-1}\left(\frac{x}{a}\right)$]
$=\frac{1}{2} \sin ^{-1}\left(\frac{8 x-9}{9}\right)$ + C
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