Problem 44 - Thomas Calculus

Question

Express your answers in terms of \(x\). \(\int \tan ^{-1} x d x\)

Step-by-Step Solution

Answer

\( \int \tan^{-1}(x) dx = x \tan^{-1}(x) - \frac{1}{2} \ln|1+x^2| + C \).

Step 1: Understand the Problem

We need to find the indefinite integral of \( \tan^{-1}(x) \) with respect to \( x \). This means we are looking for a function \( F(x) \) such that \( F'(x) = \tan^{-1}(x) \).

Step 2: Apply Integration by Parts Formula

Integration by parts is given by the formula \( \int u \, dv = uv - \int v \, du \), where we choose \( u \) and \( dv \) such that the integration becomes simpler. Let's use \( u = \tan^{-1}(x) \) and \( dv = dx \).

Step 3: Differentiate u and Integrate dv

First, find \( du \) by differentiating \( u \). The derivative \( \frac{d}{dx} \tan^{-1}(x) = \frac{1}{1 + x^2} \), so \( du = \frac{1}{1 + x^2} dx \). Next, integrate \( dv \) to find \( v \). Since \( dv = dx \), integrating gives \( v = x \).

Step 4: Substitute into Integration by Parts Formula

Substitute \( u = \tan^{-1}(x) \), \( dv = dx \), \( du = \frac{1}{1 + x^2} dx \), and \( v = x \) into the integration by parts formula: \[ \int \tan^{-1}(x) dx = x \tan^{-1}(x) - \int x \frac{1}{1+x^2} dx. \]

Step 5: Solve the Remaining Integral

The remaining integral \( \int \frac{x}{1+x^2} dx \) can be solved by substitution. Let \( w = 1 + x^2 \) so \( dw = 2x dx \). Hence, \( x dx = \frac{1}{2} dw \). Rewriting in terms of \( w \), the integral becomes \[ \int \frac{1}{2w} dw = \frac{1}{2} \ln|w| + C. \] Substitute \( w = 1 + x^2 \) back to get \( \frac{1}{2} \ln|1+x^2| + C \).

Step 6: Combine the Results

Combine the results from steps 4 and 5: \[ \int \tan^{-1}(x) dx = x \tan^{-1}(x) - \left( \frac{1}{2} \ln|1+x^2| \right) + C. \] This is the expression for the indefinite integral of \( \tan^{-1}(x) \).

Key Concepts

Indefinite Integrals

Indefinite integrals are the reverse of derivatives. They help us find a function whose derivative leads us back to the original function, like winding back the clock. When we search for an indefinite integral of a function, we are essentially solving for that mysterious original function. This is denoted as \( \int f(x) \, dx = F(x) + C \), where \( C \) is the constant of integration. It represents any constant value since the derivative of a constant is zero and doesn't affect the result.Understanding indefinite integrals is crucial because they form the basis for unlocking more complex calculus problems. By knowing that integrating is about 'undoing' differentiation, we can gradually build our skills to tackle more advanced topics like integration by parts used in the given exercise.

Substitution Method

The substitution method is a beautiful technique in calculus that helps simplify complex integrals by changing variables. We often liken it to solving a puzzle. The core idea is to substitute a part of the integral, usually one that’s making things complicated, with a simpler variable.In our exercise, we tackled the integral \( \int \frac{x}{1+x^2} \, dx \) using substitution. By letting \( w = 1 + x^2 \), we shifted the perspective of the problem. Now, differentiating \( w \) gives \( dw = 2x \, dx \), prompting us to replace \( x \, dx \) with \( \frac{1}{2} dw \). This changes the original puzzle to a simpler form: \( \int \frac{1}{2w} \, dw \). As you proceed with substitution, it's essential to fully revert back to the original variable. Once we integrate \( \frac{1}{2w} \, dw \) as \( \frac{1}{2} \ln|w| + C \) and substitute back for \( w \), we conclude the integral in terms of \( x \), simplifying our original problem. The substitution method doesn't just solve; it transforms problems, making them more manageable.

Inverse Trigonometric Functions

Inverse trigonometric functions help us find angles for which a given trigonometric value exists. They reverse the trigonometric operations, just like how roots reverse powers. The notation \( \tan^{-1}(x) \) denotes the angle whose tangent is \( x \). In calculus, these are rich grounds for integration.When dealing with integrals involving inverse trigonometric functions, it is often useful to apply techniques like integration by parts. In our exercise, we began the process by setting \( u = \tan^{-1}(x) \) due to its complexity and taking its derivative as \( du = \frac{1}{1+x^2} \, dx \). This transformation paves the way for a simpler, solution-driven method.By understanding inverse trigonometric functions within integrals, you're mastering an essential tool. Not only does it open doors to solving real-world geometric problems, but it also strengthens your grasp on calculus as a whole. Mastery of these functions ensures a solid foundation in solving intricate integrals successfully.