Question

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

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

Hint:

Whenever you see a quadratic inside a square root, try completing the square to convert into inverse trigonometric form.

Answer 1

The Correct Option is C

Concept: Use the standard form: \[ \int \frac{dx}{\sqrt{ax - bx^2}} = \sin^{-1}(\text{linear expression}) \] after completing the square.
Step 1: Rewrite the expression inside root. \[ 9x - 4x^2 = -4\left(x^2 - \frac{9}{4}x\right) \]
Step 2: Complete the square. \[ x^2 - \frac{9}{4}x = \left(x - \frac{9}{8}\right)^2 - \frac{81}{64} \] \[ 9x - 4x^2 = \frac{81}{16} - \left(x - \frac{9}{8}\right)^2 \]
Step 3: Use substitution. Let \( x - \frac{9}{8} = \frac{9}{8}\sin\theta \) \[ \Rightarrow \int \frac{dx}{\sqrt{9x - 4x^2}} = \frac{1}{2}\sin^{-1}\left(\frac{8x-9}{9}\right) + C \]