Exercise 7.4 - Question 3
Problem
$$\int \frac{1}{\sqrt{(2-x)^2 + 1}} \, dx$$
Step-by-Step Solution
Step 1: Identify the Form
The integrand has the form:
$$\frac{1}{\sqrt{u^2 + a^2}}$$
where u = (2 - x) and a = 1.
Step 2: Set Up the Substitution
Let:
$$u = 2 - x$$
Then:
$$\frac{du}{dx} = -1 \quad \Rightarrow \quad du = -dx$$
Therefore:
$$dx = -du$$
Step 3: Substitute into the Integral
Replace (2 - x) with u and dx with -du:
$$\int \frac{1}{\sqrt{(2-x)^2 + 1}} \, dx = \int \frac{1}{\sqrt{u^2 + 1}}
\cdot (-du)$$
Factor out the negative sign:
$$= -\int \frac{1}{\sqrt{u^2 + 1}} \, du$$
Step 4: Apply the Standard Formula
The standard integral is:
$$\int \frac{1}{\sqrt{u^2 + 1}} \, du = \sinh^{-1}(u) + C = \ln\left|u +
\sqrt{u^2 + 1}\right| + C$$
Applying this:
$$-\int \frac{1}{\sqrt{u^2 + 1}} \, du = -\sinh^{-1}(u) + C$$
Step 5: Substitute Back
Replace u with (2 - x):
$$= -\sinh^{-1}(2-x) + C$$
Or in logarithmic form:
$$= -\ln\left|(2-x) + \sqrt{(2-x)^2 + 1}\right| + C$$
✅ Final Answer
$$\int \frac{1}{\sqrt{(2-x)^2 + 1}} \, dx = -\sinh^{-1}(2-x) + C$$
Or equivalently:
$$= -\ln\left|(2-x) + \sqrt{(2-x)^2 + 1}\right| + C$$
Key Concepts Used
- U-Substitution: u = 2 - x leads to du = -dx
- Standard Form: ∫1/√(u² + 1) du = sinh⁻¹(u) + C
- Negative Sign: Comes from the derivative of (2 - x)