Problem 39 - Calculus Early Transcendentals

Question

Determine the following indefinite integrals. Check your work by differentiation. $$\int\left(\sec ^{2} x-1\right) d x$$

Step-by-Step Solution

Answer

Question: Find the indefinite integral of the function \(f(x) = \sec^2 x - 1\). Solution: The indefinite integral can be represented as: $$\int\left(\sec^2 x - 1\right) dx$$ Applying the step-by-step solution, we have found the indefinite integral to be: $$\int\left(\sec^2 x - 1\right) dx = \tan x + x + C$$ where C is the integration constant.

Step 1: 1. Integration of \(\sec^2 x\)

The integral of \(\sec^2 x\) can be found using its standard formula: $$\int\sec^2 x dx = \tan x + C_1$$ where \(C_1\) is the integration constant.

Step 2: 2. Integration of \(1\)

To find the integral of \(1\) with respect to \(x\), we can simply treat it as: $$\int1 dx = x + C_2$$ where \(C_2\) is another integration constant.

Step 3: 3. Combine the results

Now, let's put together the obtained results to find the integral of our initial function: $$\int\left(\sec^2 x - 1\right) dx = \tan x + C_1 + x + C_2$$ Let \(C = C_1 + C_2\), then our final result is: $$\int\left(\sec^2 x - 1\right) dx = \tan x + x + C$$

Step 4: 4. Check the result by differentiation

Now, let's differentiate our result and check if it gives us the original function: $$ \frac{d\left(\tan x + x + C\right)}{dx} = \frac{d(\tan x)}{dx} + \frac{dx}{dx} + \frac{dC}{dx} $$ Since \(\frac{d(\tan x)}{dx} = \sec^2 x\), \(\frac{dx}{dx} = 1\), and \(\frac{dC}{dx} = 0\), we get: $$ \frac{d\left(\tan x + x + C\right)}{dx} = \sec^2 x + 1 -1 = \sec^2 x - 1 $$ Since the derivative matches the initial function, our solution is correct.