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Chapter 4: Problem 42

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

Short Answer

Expert verified
Question: Determine the indefinite integral of the given function and verify your result by differentiation: $$\int \frac{\sin \theta-1}{\cos ^{2} \theta} d \theta$$ Answer: The indefinite integral of the given function is: $$\int \frac{\sin \theta-1}{\cos ^{2} \theta} d \theta = \theta - 2\tan\theta + C$$, which is verified to be correct by differentiation.

Step by step solution

01

Rewrite the integral using trigonometric identity

The given integral is: $$\int \frac{\sin \theta-1}{\cos ^{2} \theta} d \theta$$ We can rewrite \(\sin \theta\) using the identity \(\sin^2\theta+\cos^2\theta=1\) as: $$\int \frac{(\cos^2 \theta - 1) - 1}{\cos^{2} \theta} d \theta$$
02

Simplify the expression and separate the integral

Now, simplify the expression inside the integral: $$\int \frac{\cos^2\theta-2}{\cos^2\theta} d\theta = \int (1 - \frac{2}{\cos^2\theta}) d\theta$$ Then, separate the integral into two parts: $$\int 1 d\theta - 2 \int \frac{1}{\cos^2\theta} d\theta$$
03

Calculate the integrals separately

Now, solve the two integrals separately: $$\int 1 d\theta = \theta + C_1$$ For the second part, recognize that \(\int \frac{1}{\cos^2\theta} d\theta\) is the integral of the secant squared function, which is the derivative of the tangent function. Therefore, $$2 \int \frac{1}{\cos^2\theta} d\theta = 2\tan\theta + C_2$$
04

Combine the results and write the general solution

Now combine the results from Step 3: $$\theta - 2\tan\theta + C$$ where \(C = C_1 + C_2\) is the constant of integration. So, the indefinite integral of the given function is: $$\int \frac{\sin \theta-1}{\cos ^{2} \theta} d \theta = \theta - 2\tan\theta + C$$
05

Check the result by differentiation

To verify our result, we will differentiate the obtained expression: $$\frac{d}{d\theta} (\theta - 2\tan\theta + C)$$ Using simple differentiation rules, we get: $$\frac{d\theta}{d\theta} - 2\frac{d(\tan\theta)}{d\theta} = 1 - 2\sec^2\theta$$ Recall that \(\sec\theta = \frac{1}{\cos\theta}\), so \(\sec^2\theta = \frac{1}{\cos^2\theta}\). Substitute this back into the expression: $$1 - 2\frac{1}{\cos^2\theta} = \frac{\sin \theta-1}{\cos^2\theta}$$ The differentiated expression matches the original function, which confirms that our integration result is correct.

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