Question

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

• A. $\frac{1}{9}{\sin }^{-1}\left(\frac{9x-8}{8}\right)+C$
• B. $\frac{1}{2}{\sin }^{-1}\left(\frac{8x-9}{9}\right)+C$
• C. $\frac{1}{3}{\sin }^{-1}\left(\frac{9x-8}{8}\right)+C$
• D. $\frac{1}{2}{\sin }^{-1}\left(\frac{9x-8}{9}\right)+C$

Answer 1

Answer: (B)

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

$\int \frac{dx}{\sqrt{9x-4x^2}}$
$=\int \frac{1}{\sqrt{-4(x^2-\frac{9}{4}x)}}dx$
$=\int \frac{1}{-4(x^2-\frac{9}{4}x+\frac{81}{64}-\frac{81}{4})}dx$
$=\int \frac{1}{\sqrt{-4[{(x-\frac{9}{8})}^2-{\left(\frac{9}{8}\right)}^2]}}dx$
$=\frac{1}{2}\int \frac{1}{\sqrt{{\left(\frac{9}{8}\right)}^2-{(x-\frac{9}{8})}^2}}dx$
$=\frac{1}{2}\left[{\sin }^{-1}\left(\frac{x-\frac{9}{8}}{\frac{9}{8}}\right)\right]+C,(\because \int \frac{dy}{\sqrt{a^2-y^2}}={\sin }^{-1}\frac{y}{a}+C)$
$=\frac{1}{2}{\sin }^{-1}\left(\frac{8x-9}{9}\right)+C$