Exercise 7.4 - Question 4
Problem
$$\int \frac{1}{\sqrt{9 - 25x^2}} \, dx$$
Step-by-Step Solution
Step 1: Identify the Form
This is of the form:
$$\frac{1}{\sqrt{a^2 - u^2}}$$
We need to rewrite 9 - 25x² in this form.
Step 2: Rewrite the Expression
Factor out constants:
$$9 - 25x^2 = 9\left(1 - \frac{25x^2}{9}\right) = 9\left(1 -
\left(\frac{5x}{3}\right)^2\right)$$
So:
$$\sqrt{9 - 25x^2} = 3\sqrt{1 - \left(\frac{5x}{3}\right)^2}$$
Step 3: Set Up Substitution
Let:
$$u = \frac{5x}{3}$$
Then:
$$du = \frac{5}{3}dx \quad \Rightarrow \quad dx = \frac{3}{5}du$$
Step 4: Substitute
$$\int \frac{1}{3\sqrt{1 - u^2}} \cdot \frac{3}{5}du = \frac{1}{5}\int
\frac{1}{\sqrt{1 - u^2}} \, du$$
Step 5: Apply Standard Formula
The standard formula is:
$$\int \frac{1}{\sqrt{1 - u^2}} \, du = \sin^{-1}(u) + C$$
Therefore:
$$\frac{1}{5}\int \frac{1}{\sqrt{1 - u^2}} \, du = \frac{1}{5}\sin^{-1}(u) +
C$$
Step 6: Substitute Back
Replace u with 5x/3:
$$= \frac{1}{5}\sin^{-1}\left(\frac{5x}{3}\right) + C$$
✅ Final Answer
$$\int \frac{1}{\sqrt{9 - 25x^2}} \, dx =
\frac{1}{5}\sin^{-1}\left(\frac{5x}{3}\right) + C$$
Key Concepts Used
- Standard Form: ∫1/√(a² - u²) du = sin⁻¹(u/a) + C
- Factoring: Extracting constants to match standard form