Problem 45

Question

Express your answers in terms of $x$. $\int \sec ^{-1} x d x$

Step-by-Step Solution

Verified

Answer

$ \int \sec^{-1} x \, dx = x \sec^{-1} x - \ln |x + \sqrt{x^2 - 1}| + C $

1Step 1: Identify the Integral Type

The integral given is $ \int \sec^{-1} x \, dx $. This is an inverse trigonometric function integral, which often involves integration techniques like integration by parts.

2Step 2: Choose Parts for Integration by Parts

Use integration by parts formula $ \int u \, dv = uv - \int v \, du $. For $ u $, choose $ \sec^{-1} x $ and for $ dv $, choose $ dx $.

3Step 3: Differentiate and Integrate

Differentiate $ u $: $ du = \frac{1}{|x| \sqrt{x^2 - 1}} \, dx $ and integrate $ dv $: $ v = x $.

4Step 4: Apply Integration by Parts Formula

Substitute into integration by parts formula: $ \int \sec^{-1} x \, dx = x \sec^{-1} x - \int x \frac{1}{|x| \sqrt{x^2 - 1}} \, dx $.

5Step 5: Simplify and Integrate Second Part

Simplify $ \int x \frac{1}{|x| \sqrt{x^2 - 1}} \, dx $ to $ \int \frac{1}{\sqrt{x^2 - 1}} \, dx $, which is $ \ln |x + \sqrt{x^2 - 1}| + C $.

6Step 6: Final Integration Result

Combine the results: $ x \sec^{-1} x - \ln |x + \sqrt{x^2 - 1}| + C $. This is the final expression for $ \int \sec^{-1} x \, dx $.

Key Concepts

Inverse Trigonometric FunctionsIntegral CalculusIntegration Techniques

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverse operations of the basic trigonometric functions. They help us determine the angle measure when the value of the trigonometric function is known. For example, the function $ \sec^{-1} x $ is the inverse of the secant function $ \sec(\theta) $, where it gives the angle $ \theta $ for a given secant value $ x $. These functions are crucial in calculus because they appear in various integral and derivative calculations, making them key players in solving more complex problems.

There are six primary inverse trigonometric functions: $ \sin^{-1} x $, $ \cos^{-1} x $, $ \tan^{-1} x $, $ \csc^{-1} x $, $ \sec^{-1} x $, and $ \cot^{-1} x $.
Each function has specific domains and ranges, carefully chosen to ensure they are valid inverse functions.
The function $ \sec^{-1} x $ is defined for $ x \le -1 $ or $ x \ge 1 $.

Along with these properties, understanding their derivatives and integrals contributes to the grasp of integral calculus.

Integral Calculus

Integral calculus is the branch of calculus that deals with finding the integral of functions, which is essentially the reverse process of differentiation. Integrals are used to calculate areas under curves, volumes, and other quantities that accumulate continuously.

The definite integral calculates the net area between the function and the x-axis over a specific interval $[a, b]$.
The indefinite integral, such as $ \int \sec^{-1} x \, dx $, finds the general form of the antiderivative, expressed along with a constant $ C $ that represents all possible vertical translations of the function.
Techniques like substitution and integration by parts are applied when direct integration is not feasible.

In this exercise, the goal is to find the indefinite integral of $ \sec^{-1} x $, requiring the use of a specific method due to the complexity of the inverse function involved.

Integration Techniques

When facing complex integrals, such as those involving inverse trigonometric functions, specific integration techniques simplify the process. One key technique is **integration by parts**, based on the product rule of differentiation and designed to handle products of functions.
Here’s how integration by parts works:

Choose a part of the integral to differentiate ($ u $) and another part to integrate ($ dv $).
Differentiate $ u $ to find $ du $ and integrate $ dv $ to find $ v $.
Substitute in the formula: $ \int u \, dv = uv - \int v \, du $.

In the exercise $ \int \sec^{-1} x \cdot 1 \, dx $, $ \sec^{-1} x $ was chosen as $ u $, while $ dx $ became $ dv $. This smart choice simplifies the process by turning a challenging integral into two simpler ones to solve step by step.
With practice, the choice of $ u $ and $ dv $ becomes more intuitive, helping you tackle a broad range of integration problems.

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