Question

\(\int \dfrac{dx}{\sqrt{9x - 4x^2}}\) equals:
\( (A) \dfrac{1}{9} \sin^{-1} \left(\dfrac{9x -8}{8} \right ) + C\)
\( (B) \dfrac{1}{2} \sin^{-1} \left(\dfrac{8x - 9}{9} \right ) + C\)
\((C) \dfrac{1}{3} \sin^{-1} \left(\dfrac{9x - 8}{8} \right ) + C\)
\( (D) \dfrac{1}{2} \sin^{-1} \left(\dfrac{9x -8}{8} \right ) + C\)

Answer 1

\(\int \dfrac{dx}{\sqrt{9x - 4x}^2}\)
\( = \int \dfrac{dx}{\sqrt{-4 \left(x^2 - \dfrac{9}{4}x \right )}}\)
\( = \dfrac{1}{2} \int \dfrac{dx}{\sqrt{-\left(x^2 - \dfrac{9}{4} x + \left(\dfrac{9}{8} \right )^2 - \left(\dfrac{9}{8} \right )^2 \right )}}\)
\( = \dfrac{1}{2} \int \dfrac{dx}{\sqrt{-\left[\left(x - \dfrac{9}{8} \right )^2 - \left(\dfrac{9}{8} \right )^2 \right ]}}\)
\( = \dfrac{1}{2} \int \dfrac{dx}{\sqrt{\left(\dfrac{9}{8} \right )^2} - \left(x - \dfrac{9}{8} \right )^2}\)
\( = \dfrac{1}{2} \left [\sin^{-1} \left(\dfrac{x - \dfrac{9}{8}}{\dfrac{9}{8}} \right ) \right ] + C\)
\( \left[\because \int \dfrac{1}{\sqrt{a^2 - x^2}}dx = \sin^{-1} \dfrac{x}{a} + C \right ]\)
\( = \dfrac{1}{2} \sin^{-1} \left(\dfrac{8x -9}{9} \right ) + C\)
Where C is constant of integration
\(\therefore\) Option B is correct.