Problem 43 - Calculus for Biology and Medicine

Question

$$ \text { In Problems } , \text { evaluate each definite integral. } $$ $$ \int_{0}^{1} \tan ^{-1} x d x $$

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{\pi}{4} - \frac{1}{2} \ln 2\).

Step 1: Understand the Problem

We need to evaluate the definite integral of the inverse tangent function, \(\tan^{-1}(x)\), from 0 to 1. This is written mathematically as \(\int_{0}^{1} \tan^{-1} x\, dx\).

Step 2: Choose the Integration Technique

Since the integrand \(\tan^{-1} x\) does not have an elementary antiderivative, we will use integration by parts, which is suitable for this kind of problem. Integration by parts is given by:\[\int u\, dv = uv - \int v\, du\]We need to choose \(u\) and \(dv\).

Step 3: Selecting Functions for Integration by Parts

Let \(u = \tan^{-1} x\), which means \(du = \frac{1}{1+x^2}\, dx\). Let \(dv = dx\), then \(v = \int dx = x\).

Step 4: Apply Integration by Parts Formula

Substitute the chosen functions into the integration by parts formula:\[\int \tan^{-1} x\, dx = x \tan^{-1} x - \int x \left( \frac{1}{1+x^2} \right) dx\]We now need to solve the remaining integral.

Step 5: Solve the Remaining Integral

The integral \(\int \frac{x}{1+x^2}\, dx\) can be solved using a simple substitution. Set \(w = 1 + x^2\), then \(dw = 2x\, dx\) or \(x\, dx = \frac{1}{2} dw\). This transforms the integral into:\[\int \frac{x}{1+x^2} dx = \frac{1}{2}\int \frac{1}{w} dw = \frac{1}{2} \ln |w| + C\]Substitute back for \(w\):\[\frac{1}{2} \ln |1+x^2|\]

Step 6: Combine Results and Evaluate Definite Integral

Reintegrate the result back to the original expression and evaluate from 0 to 1:\[\begin{align*}\int_{0}^{1} \tan^{-1} x\, dx = \left[ x \tan^{-1} x - \frac{1}{2} \ln |1+x^2| \right]_0^1 = \left[ 1 \cdot \frac{\pi}{4} - \frac{1}{2} \ln 2 \right] - \left[ 0 - \frac{1}{2} \ln 1 \right]\end{align*}\]Thus, the integral evaluates to \(\frac{\pi}{4} - \frac{1}{2} \ln 2\).

Key Concepts

Integration by Parts

Integration by parts is a powerful technique used in calculus to solve integrals that are products of functions. Even if this seems complex, it often simplifies the integration process. The integration by parts formula is a clever approach that leverages the product rule of differentiation in reverse. The core formula is:

• \( \int u \, dv = uv - \int v \, du \)

The trick lies in choosing the right function for \( u \) and \( dv \). Typically, we choose \( u \) as a function that gets simpler upon differentiation, and \( dv \) as a function that is easy to integrate. In our problem, for example:

• Let \( u = \tan^{-1} x \), hence \( du = \frac{1}{1+x^2} \, dx \).
• Let \( dv = dx \), hence \( v = x \).

By substituting these into the formula, the integral can be slowly broken down into more manageable parts. Always remember that solving an integral by parts often doesn't finish in one step. You may need to apply the technique repeatedly or use additional methods to handle what's left.

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverse operations of regular trigonometric functions, like sine or cosine. They are crucial when dealing with angles that produce given trigonometric values. In our integral:

• \( \tan^{-1}(x) \) is the inverse tangent function.

It helps determine the angle whose tangent value is \( x \). Inverse trigonometric functions take ordinary numbers and return angles in specified ranges. This becomes particularly handy in integrals, as such functions don't have elementary (simple) antiderivatives. Consequently, methods like integration by parts offer a way to work around this and evaluate the integral efficiently.
Understanding the domains and ranges of these functions can simplify calculations substantially. For instance, knowing \( \tan^{-1}(x) \) returns values between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\), helps in evaluating integrals with definite limits.

Definite Integrals

Definite integrals extend the concept of antiderivatives to calculating actual values over intervals. When you evaluate a definite integral, you calculate the net area under a curve between two points, which offers tremendous applications in physics and engineering. The notation:

• \( \int_{a}^{b} f(x) \, dx \)

refers to the integral of the function \( f(x) \) from the lower limit \( a \) to the upper limit \( b \). In our problem, the definite integral \( \int_{0}^{1} \tan^{-1} x \, dx \) requires evaluating the antiderivative at these limits and finding their difference.

After solving the indefinite part of the integral: \( x \tan^{-1} x - \frac{1}{2} \ln |1+x^2| \), which we computed using integration by parts, the final step involves plugging in these boundary values (0 and 1) into the antiderivative.

• This process gives the specific area or accumulation of change from one point to another, which is what definite integrals beautifully represent.