Problem 44 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\pi}{8} - \frac{1}{4} + \frac{1}{4} \ln 2 \).

Step 1: Identify the Function and Integral

We need to evaluate the definite integral \( \int_{0}^{1} x \tan^{-1} x \, dx \). This integral involves a product of functions: \( x \) and \( \tan^{-1} x \).

Step 2: Choose a Method for Integration

Since the integral is a product of two functions, we'll use integration by parts. Recall the formula for integration by parts: \( \int u \, dv = uv - \int v \, du \). We will choose \( u = \tan^{-1}x \) and \( dv = x \, dx \).

Step 3: Differentiate and Integrate Selected Parts

Differentiate \( u \) and integrate \( dv \):- \( u = \tan^{-1}x \), hence \( du = \frac{1}{1+x^2} \, dx \).- \( dv = x \, dx \), hence \( v = \frac{x^2}{2} \).

Step 4: Apply the Integration by Parts Formula

Substitute the values into the integration by parts formula:\[\int x \tan^{-1} x \, dx = \left[ \tan^{-1} x \cdot \frac{x^2}{2} \right]_0^1 - \int_0^1 \frac{x^2}{2(1+x^2)} \, dx\]

Step 5: Integrate the Remaining Integral

Simplify and integrate the remaining integral:\[\int_0^1 \frac{x^2}{2(1+x^2)} \, dx = \frac{1}{2} \int_0^1 \frac{x^2}{1+x^2} \, dx \]This integral can be simplified by the substitution method: let \( w = 1 + x^2 \), then \( dw = 2x \, dx \), and the limits of integration change as follows: when \( x = 0, w = 1 \); when \( x = 1, w = 2 \).Substitute and integrate:\[\frac{1}{2} \int_1^2 \frac{w-1}{w} \cdot \frac{1}{2} \, dw = \frac{1}{4} \int_1^2 1 - \frac{1}{4} \int_1^2 \frac{1}{w} \, dw \]\[= \frac{1}{4} \left[ w \right]_1^2 - \frac{1}{4} \left[ \ln|w| \right]_1^2 = \frac{1}{4} (2 - 1) - \frac{1}{4} (\ln 2 - \ln 1) \]\[= \frac{1}{4} - \frac{1}{4} \ln 2\]

Step 6: Evaluate at the Boundaries and Combine Results

Now substitute back into the integration by parts result and evaluate from 0 to 1:- Evaluate \( \left[ \tan^{-1} x \cdot \frac{x^2}{2} \right]_0^1 = \tan^{-1} 1 \cdot \frac{1^2}{2} - \tan^{-1} 0 \cdot \frac{0^2}{2} = \frac{\pi}{4} \times \frac{1}{2} = \frac{\pi}{8} \).Combine with the result from Step 5:\[\frac{\pi}{8} - \left( \frac{1}{4} - \frac{1}{4} \ln 2 \right) = \frac{\pi}{8} - \frac{1}{4} + \frac{1}{4} \ln 2\]

Step 7: Simplify the Result

Refine the expression to get the final answer:\[\frac{\pi}{8} - \frac{1}{4} + \frac{1}{4} \ln 2\]This is the evaluated result of the definite integral \( \int_0^1 x \tan^{-1} x \, dx \).

Key Concepts

Integration by Parts

Integration by parts is a powerful technique for calculating the integral of a product of two functions. This method is particularly useful when one of the functions becomes simpler when differentiated, while the other can be easily integrated. The integration by parts formula is given by: \[ \int u \, dv = uv - \int v \, du \] To use this method, you first need to identify your parts: select which function will be \( u \) (to differentiate) and which will be \( dv \) (to integrate). Here's how to decide:

• Choose \( u \) such that \( du \) (its derivative) is simpler than \( u \). Common choices for \( u \) include inverse trigonometric, logarithmic, and algebraic functions.
• Choose \( dv \) so that it can be easily integrated to find \( v \). Functions such as polynomials and exponential functions are often good choices for \( dv \).

In the problem \( \int_{0}^{1} x \tan^{-1} x \, dx \), we selected \( u = \tan^{-1}x \) because its derivative \( du = \frac{1}{1+x^2} \, dx \) simplifies the expression considerably. Meanwhile, \( dv = x \, dx \) integrates to \( v = \frac{x^2}{2} \). After selecting these parts, substitute them into the formula and simplify to find your answer.

Substitution Method

The substitution method is another important technique used in integration, particularly useful when an integral contains a composite function. The goal of substitution is to simplify the integral by changing variables. The general idea is to let \( u \) be a function inside the integral whose derivative \( du \) will replace part of the integrand, thus simplifying the integration process. To effectively apply the substitution method, follow these steps:

• Identify the inner function of the composite function to use as your substitution variable \( u \).
• Calculate \( du \), the differential of \( u \), and solve for \( dx \) in terms of \( du \).
• Substitute \( u \) and \( du \) into the integral, replacing all instances of the original variable.
• Integrate with respect to \( u \), and finally, substitute back the original variable in terms of \( u \) to complete the integration.

In the example given, to simplify \( \int \frac{x^2}{1+x^2} \, dx \), we set \( w = 1 + x^2 \). This affects the integration limits and transforms the integral into something more straightforward, like \( \int \frac{w-1}{w} \, dw \) after substitution, making it easier to solve.

Trigonometric Integration

Trigonometric integration involves the integration of expressions containing trigonometric functions, and typically includes methods specific to these types of functions. These techniques often involve identities and substitutions to simplify the integration process. In some integrals, inverse trigonometric functions appear, such as \( \tan^{-1}(x) \) found in the original problem. When dealing with trigonometric functions:

• Consider using trigonometric identities to simplify expressions. This can include identities like \( \sin^2(x) + \cos^2(x) = 1 \).
• If the integrand involves a mix of algebraic terms and trig functions, substitution might help, especially transforming polar to rectangular coordinates or vice versa.
• Inverse trigonometric functions, such as \( \tan^{-1}(x) \), serve commonly in integration problems. Finding the derivative of inverse trigonometric functions is often part of integration by parts or substitution techniques.

In definite integrals, like \( \int_{0}^{1} x \tan^{-1} x \, dx \), using trigonometric identities or inverses aids in transforming the integrand into a more feasible form for integration.