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Chapter 4: Problem 39

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

Short Answer

Expert verified
Question: Find the indefinite integral of the function $$\int(\sec^2x-1) dx$$. Answer: The indefinite integral of the function $$\int(\sec^2x-1) dx$$ is $$\tan{x} - x + C$$, where C is the constant of integration.

Step by step solution

01

Integrate the function

To integrate the function, we will integrate each term separately: $$\int\left(\sec ^{2} x-1\right) d x = \int \sec^2{x} dx - \int 1 dx$$
02

Integrate $$\sec^2{x}$$

The antiderivative of $$\sec^2{x}$$ is well-known to be $$\tan x$$. Therefore, $$\int \sec^2{x} dx = \tan x + C_1$$, where $$C_1$$ is the constant of integration.
03

Integrate $$1$$

The antiderivative of $$1$$ is $$x$$. Therefore, $$\int 1 dx = x + C_2$$ where $$C_2$$ is another constant of integration.
04

Combine the integrals

Now, combine the antiderivatives from Steps 2 and 3: $$\int\left(\sec ^{2} x-1\right) d x = \tan x + C_1 - x - C_2$$ Combine the constants $$C_1$$ and $$C_2$$ into a single constant, $$C$$: $$\int\left(\sec ^{2} x-1\right) d x = \tan x - x + C$$
05

Check the result by differentiation

Differentiate the result with respect to $$x$$ using the chain rule: $$\frac{d}{dx}(\tan x - x) = \sec^2{x} - 1$$ This matches the given function, so our result is correct. Final answer: $$\int\left(\sec ^{2} x-1\right) d x = \tan x - x + C$$

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