# Show that

Question:

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

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

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

C. $\frac{1}{3} \sin ^{-1}\left(\frac{9 x-8}{8}\right)+\mathrm{C}$

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

Solution:

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

$=\int \frac{1}{\sqrt{-4\left(x^{2}-\frac{9}{4} x\right)}} d x$

$=\int \frac{1}{-4\left(x^{2}-\frac{9}{4} x+\frac{81}{64}-\frac{81}{64}\right)^{d x}}$

$=\int \frac{1}{\left.\sqrt{-4\left(\left(x-\frac{9}{8}\right)^{2}-\left(\frac{9}{8}\right)^{2}\right.}\right]} d x$

$=\frac{1}{2} \int \frac{1}{\sqrt{\left(\frac{9}{8}\right)^{2}-\left(x-\frac{9}{8}\right)^{2}}} d x$

$=\int \frac{1}{\left.\sqrt{-4\left[\left(x-\frac{9}{8}\right)^{2}-\left(\frac{9}{8}\right)^{2}\right.}\right]} d x$

$=\frac{1}{2}\left[\sin ^{-1}\left(\frac{x-\frac{9}{8}}{\frac{9}{8}}\right)\right]+\mathrm{C}$     $\left(\int \frac{d y}{\sqrt{a^{2}-y^{2}}}=\sin ^{-1} \frac{y}{a}+\mathrm{C}\right)$

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

Hence, the correct answer is B.