Question:
\(∫\frac{dx}{\sqrt{9x-4x^2}}\) equals
\(∫\frac{dx}{\sqrt{9x-4x^2}}\) equals
Updated On: Sep 15, 2023
\(\frac{1}{9}sin^{-1}(\frac{9x-8}{8})+C\)
\(\frac{1}{2}sin^{-1}(\frac{8x-9}{9})+C\)
\(\frac{1}{3}sin^{-1}(\frac{9x-8}{8})+C\)
\(\frac{1}{2}sin^{-1}(\frac{9x-8}{9})+C\)
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The Correct Option is B
Solution and Explanation
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.
\(∫\frac{dx}{\sqrt{9x-4x^2}}\)
\(= ∫\frac{1}{\sqrt{-4(x^2-\frac{9}{4x})}}dx\)
\(= ∫\frac{1}{-4(x^2-\frac{9}{4}x+\frac{81}{64}-\frac{81}{64})}dx\)
\(= ∫\frac{1}{\sqrt{-4[(x-\frac{9}{8})^2-(\frac{9}{8})^2}}] dx\)
\(=\frac{1}{2} ∫\frac{1}{\sqrt{(\frac{9}{8})^2 - (x-\frac{9}{8})^2}} dx\)
\(=\frac{1}{2}[sin^{-1}\bigg(\frac{x-\frac{9}{8}}{\frac{9}{8}}\bigg)]+C\) \(( ∫\frac{dy}{\sqrt{a^2-y^2}} = sin^{-1} \frac{y}{a}+C)\)
\(= \frac{1}{2}sin^{-1}(\frac{8x-9}{9})+C\)
Hence, the correct Answer is B.
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- ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
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