Question

\(∫\frac{dx}{\sqrt{9x-4x^2}}\) equals

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

Answer 1

The Correct Option is B

The correct answer is B: \(= \frac{1}{2}sin^{-1}(\frac{8x-9}{9})+C\)
\(∫\frac{dx}{\sqrt{9x-4x^2}}\)
\(= ∫\frac{1}{\sqrt{-4(x^2-\frac{9}{4x})}}dx\)
\(= ∫\frac{1}{-4(x^2-\frac{9}{4}x+\frac{81}{64}-\frac{81}{64})}dx\)
\(= ∫\frac{1}{\sqrt{-4[(x-\frac{9}{8})^2-(\frac{9}{8})^2}}] dx\)
\(=\frac{1}{2} ∫\frac{1}{\sqrt{(\frac{9}{8})^2 - (x-\frac{9}{8})^2}} dx\)
\(=\frac{1}{2}[sin^{-1}\bigg(\frac{x-\frac{9}{8}}{\frac{9}{8}}\bigg)]+C\)                              \(( ∫\frac{dy}{\sqrt{a^2-y^2}} = sin^{-1} \frac{y}{a}+C)\)
\(= \frac{1}{2}sin^{-1}(\frac{8x-9}{9})+C\)
Hence, the correct Answer is B.