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GUJCET · Maths · Integrals

\(\int \frac{dx}{\sqrt{9x - 4x^2}} =\) ___________ + C

  1. A \(\frac{1}{2} \sin^{-1} \left(\frac{8x-9}{9}\right)\)
  2. B \(\frac{1}{9} \sin^{-1} \left(\frac{9x-8}{8}\right)\)
  3. C \(\frac{1}{3} \sin^{-1} \left(\frac{9x-8}{8}\right)\)
  4. D \(\frac{1}{2} \sin^{-1} \left(\frac{9x-8}{9}\right)\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

\( \int \frac{dx}{\sqrt{9x - 4x^2}} = \int \frac{dx}{\sqrt{\frac{81}{16} - 4\left(x - \frac{9}{8}\right)^2}} \) \( = \frac{1}{2} \int \frac{dx}{\sqrt{\frac{81}{64} - \left(x - \frac{9}{8}\right)^2}} \)