Problem 414 - Calculus Volume 1

Question

In the following exercises, compute the antiderivative using appropriate substitutions. $$\int \frac{t \tan ^{-1}\left(t^{2}\right)}{1+t^{4}} d t$$

Step-by-Step Solution

Answer

The antiderivative is \( \frac{1}{2} \left( t^2 \tan^{-1}(t^2) - \frac{1}{2} \ln(1+t^4) \right) + C \).

Step 1: Identify the substitution

In the integral \( \int \frac{t \tan^{-1}(t^2)}{1+t^4} \, dt \), notice that the expression \( t^2 \) is inside the \( \tan^{-1} \) and also in the denominator in the form \( 1 + t^4 = 1 + (t^2)^2 \). We use the substitution \( u = t^2 \). Then, \( du = 2t \, dt \).

Step 2: Rewrite the integral in terms of u

Given \( u = t^2 \), \( du = 2t \, dt \) which implies \( t \, dt = \frac{1}{2} du \). Substitute these into the integral. The integral becomes \( \int \frac{\tan^{-1}(u)}{1+u^2} \frac{1}{2} \, du \), or \( \frac{1}{2} \int \frac{\tan^{-1}(u)}{1+u^2} \, du \).

Step 3: Determine a suitable integration technique

The integral \( \frac{1}{2} \int \frac{\tan^{-1}(u)}{1+u^2} \, du \) is solved using integration by parts. Choose \( v = \tan^{-1}(u) \) and \( dw = \frac{1}{1+u^2} \, du \). Differentiating and integrating gives \( dv = \frac{1}{1+u^2} \, du \) and \( w = u \), respectively.

Step 4: Apply integration by parts formula

The integration by parts formula is \( \int v \, dw = vw - \int w \, dv \). Substitute the expressions for \( v \) and \( w \): \[ \frac{1}{2} \int \tan^{-1}(u) \cdot \frac{1}{1+u^2} \, du = \frac{1}{2} \left( u \tan^{-1}(u) - \int u \cdot \frac{1}{1+u^2} \, du \right) \].

Step 5: Simplify the integral of remaining term

The remaining integral \( \int u \cdot \frac{1}{1+u^2} \, du \) simplifies easily since \( \frac{d}{du}(\ln(1+u^2)) = \frac{2u}{1+u^2} \), implying \( \int u \cdot \frac{1}{1+u^2} \, du = \frac{1}{2} \ln(1+u^2) + C \).

Step 6: Final assembly and back substitution

Replace back into \( u = t^2 \) and substitute everything to get: \[ \frac{1}{2} \left( t^2 \tan^{-1}(t^2) - \frac{1}{2} \ln(1+t^4) \right) + C \], where \( C \) is the integration constant.

Key Concepts

Substitution Method

The substitution method, often referred to as "u-substitution," is a powerful technique for finding antiderivatives, particularly in integrals where a direct calculation is not obvious. This method involves substituting part of the integral with a new variable, typically simplifying the expression to a more familiar form.

To use substitution effectively:

• Identify a part of the integrand (the function inside the integral) that you can replace with a single variable, often denoted as \( u \).
• Calculate the differential of \( u \), \( du \), and express \( dt \) in terms of \( du \).
• Rewrite the integral in terms of \( u \) and \( du \), which often results in a simpler integral.
• Once integrated, substitute back the original variable to complete the solution.

In the exercise, \( t^2 \) was replaced with \( u \), making the integral simpler to handle through substitution. The simplified form allowed the application of further techniques, like integration by parts.

Integration by Parts

Integration by parts is another method for solving integrals and is particularly useful when the integrand is a product of two functions. This technique is derived from the product rule for differentiation and provides a formula to help break down complex integrals.The formula for integration by parts is:\[\int u \, dv = uv - \int v \, du\]Steps to apply this method include:

• Identify two components in the integrand: choose one as \( u \) and the other as \( dv \).
• Differentiate \( u \) to get \( du \), and integrate \( dv \) to obtain \( v \).
• Substitute into the integration by parts formula.
• Simplify and solve any remaining integrals as needed.

In the given exercise, \( u = \tan^{-1}(u) \) and \( dv = \frac{1}{1+u^2} \, du \) were used, which leveraged the derivative of inverse trigonometric functions to simplify the integration process further.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as \( \tan^{-1}(x) \), frequently appear in calculus, particularly in integrals where they may result in more manageable computations or signal specific integration techniques. These functions are the inverses of basic trigonometric functions and have well-defined derivatives, which are key when solving integrals involving these functions.For \( \tan^{-1}(x) \), the derivative is:\[\frac{d}{dx}[\tan^{-1}(x)] = \frac{1}{1+x^2}\]These derivatives facilitate integration by parts as seen in the exercise. Recognizing these derivatives in the integrand can greatly assist in choosing the right technique for integration.

In integrals, using inverse trig functions can sometimes simplify otherwise intricate problems, making them pivotal in upper-level calculus homework and exercises.

Definite and Indefinite Integrals

Understanding the difference between definite and indefinite integrals is crucial in calculus. An indefinite integral represents a family of functions and is denoted by:\[\int f(x) \, dx = F(x) + C\]where \( C \) is the constant of integration. This is what we obtained in the exercise, resulting in a general antiderivative without specific limits of integration.Definite integrals, on the other hand, compute the net area under a curve between two points and are represented as:\[\int_{a}^{b} f(x) \, dx\]where \( a \) and \( b \) are the limits of integration.

• Indefinite integrals yield a general solution for a function.
• Definite integrals determine a specific numerical value, representing the area under a curve.

In calculus exercises, recognizing which type of integral to solve is vital, and when using techniques like substitution or integration by parts, one often starts with indefinite integrals to ease calculations.