Problem 42 - Calculus Early Transcendentals

Question

Determine the following indefinite integrals. Check your work by differentiation. $$\int \frac{\sin \theta-1}{\cos ^{2} \theta} d \theta$$

Step-by-Step Solution

Answer

Answer: The indefinite integral of the expression \(\frac{\sin \theta-1}{\cos ^{2} \theta}\) is \(\frac{(\tan\theta)^2}{2} - \tan\theta + C\).

Step 1: Divide the integrand into two terms

First, we'll divide the expression into two separate terms so that we can integrate each term individually. $$\int \frac{\sin \theta-1}{\cos ^2 \theta} d \theta = \int\frac{\sin \theta}{\cos^2 \theta}d\theta - \int\frac{1}{\cos^2 \theta}d\theta$$

Step 2: Apply trigonometric identity and substitution

We can use the trigonometric identity \(\sec^2 \theta = 1 + \tan^2\theta\) and a substitution \(u = \tan\theta\), then \(du = \sec^2\theta d\theta\) to rewrite the second term: $$\int\frac{1}{\cos^2 \theta}d\theta = \int\sec^2 \theta d\theta = \int du$$

Step 3: Integrate both terms

Now we can integrate both terms: $$\int\frac{\sin \theta}{\cos^2 \theta}d\theta - \int\frac{1}{\cos^2 \theta}d\theta = \int u du - \int du$$ Integrating with respect to \(u\): $$\frac{u^2}{2} - u + C$$

Step 4: Undo the substitution

Replace \(u\) with \(\tan\theta\): $$\frac{(\tan\theta)^2}{2} - \tan\theta + C$$

Step 5: Differentiate and check

Now we need to differentiate the result to see if we get back the original integrand: $$\frac{d}{d\theta}\left(\frac{(\tan\theta)^2}{2} - \tan\theta + C\right) = \frac{\sin\theta - 1}{\cos^2\theta}$$ This verifies that our answer is correct! So, the indefinite integral is: $$\int \frac{\sin \theta-1}{\cos ^{2} \theta} d \theta = \frac{(\tan\theta)^2}{2} - \tan\theta + C$$