Exercise 7.4 - Question 25 (MCQ)
Problem
$$\int \frac{dx}{\sqrt{9x - 4x^2}} \text{ equals}$$
(A) (1/9)sin⁻¹((9x-8)/8) + C
(B) (1/2)sin⁻¹((8x-9)/9) + C
(C) (1/3)sin⁻¹((9x-8)/8) + C
(D) (1/2)sin⁻¹((9x-8)/9) + C
Step-by-Step Solution
Step 1: Factor Out
$$9x - 4x^2 = -4\left(x^2 - \frac{9x}{4}\right) = -4\left[\left(x -
\frac{9}{8}\right)^2 - \frac{81}{64}\right]$$
$$= \frac{81}{16} - 4\left(x - \frac{9}{8}\right)^2 =
\left(\frac{9}{4}\right)^2 - (2x - \frac{9}{4})^2$$
Step 2: Simplify
$$\sqrt{9x - 4x^2} = \sqrt{\frac{81}{16} - \left(x - \frac{9}{8}\right)^2
\cdot 4}$$
Step 3: Substitute and Apply Formula
Let u = x - 9/8, the integral becomes:
$$\int \frac{1}{\sqrt{(9/4)^2 - (2u)^2}} \, du =
\frac{1}{2}\sin^{-1}\left(\frac{2u}{9/4}\right) + C$$
$$= \frac{1}{2}\sin^{-1}\left(\frac{8x - 9}{9}\right) + C$$
✅ Answer: (B) (1/2)sin⁻¹((8x-9)/9) + C
$$\int \frac{dx}{\sqrt{9x - 4x^2}} =
\frac{1}{2}\sin^{-1}\left(\frac{8x-9}{9}\right) + C$$