Exercise 7.4 - Question 14
Problem
$$\int \frac{1}{\sqrt{8 + 3x - x^2}} \, dx$$
Step-by-Step Solution
Step 1: Complete the Square
$$8 + 3x - x^2 = -\left(x^2 - 3x - 8\right) = -\left[\left(x -
\frac{3}{2}\right)^2 - \frac{9}{4} - 8\right]$$
$$= -\left(x - \frac{3}{2}\right)^2 + \frac{41}{4} = \frac{41}{4} - \left(x
- \frac{3}{2}\right)^2$$
Step 2: Rewrite
$$\int \frac{1}{\sqrt{(\frac{\sqrt{41}}{2})^2 - (x - \frac{3}{2})^2}} \,
dx$$
Step 3: Apply Formula
Using ∫1/√(a² - u²) du = sin⁻¹(u/a):
$$= \sin^{-1}\left(\frac{x - 3/2}{\sqrt{41}/2}\right) + C =
\sin^{-1}\left(\frac{2x - 3}{\sqrt{41}}\right) + C$$
✅ Final Answer
$$\int \frac{1}{\sqrt{8 + 3x - x^2}} \, dx = \sin^{-1}\left(\frac{2x -
3}{\sqrt{41}}\right) + C$$