Problem 39 - Essential Calculus

Question

Evaluate the definite integral. Use a graphing utility to verify your result. $$ \int_{1}^{2} \frac{1-\cos \theta}{\theta-\sin \theta} d \theta $$

Step-by-Step Solution

Answer

The definite integral equals \(\ln \left|\frac{2 - \sin(2)}{1 - \sin(1)}\right|\)

Step 1: Identify the Substitution

Observe the integrand and try to identify a function \(u(\theta)\) and its derivative \(du/d\theta\). Here, choose \(u(\theta) = \theta - \sin(\theta)\) because its derivative, which is \(du/d\theta = 1-\cos(\theta)\), is present in the integrand.

Step 2: Substitute and Simplify

Express the integrand in terms of \(u\). Replace \(\theta - \sin(\theta)\) by \(u\) and \(1 - \cos(\theta) d\theta\) by \(du\). The integrand simplifies to \(1/u\). Simultaneously, transform the limits of integration according to the chosen \(u(\theta)\). When \(\theta = 1\), \(u = 1 - \sin(1)\) and when \(\theta = 2\), \(u = 2 - \sin(2)\). Therefore, the integral turns into: \(\int_{1 - \sin(1)}^{2 - \sin(2)} du / u\).

Step 3: Solve the Integral

The integral of \(1/u\) with respect to \(u\) is \(\ln |u|\), so the antiderivative of \(du/u\) is \(\ln |u|\). Therefore, we get the result as \(\ln |u| \Big|_{1 - \sin(1)}^{2 - \sin(2)}\).

Step 4: Apply the Fundamental Theorem of Calculus

Use the Fundamental Theorem of Calculus to evaluate the definite integral. Subtract the value of the antiderivative at the lower limit of integration from the value at the upper limit. The result is \(\ln |2 - \sin(2)| - \ln |1 - \sin(1)|\), which simplifies to: \(\ln \left|\frac{2 - \sin(2)}{1 - \sin(1)}\right|\).