Problem 25 - AP Calculus with 8 Practice Tests 15th

Question

\(\int t \cos \left(4 t^{2}\right) d t=\) (A) \(\frac{1}{8} \sin \left(4 t^{2}\right)+C\) (B) \(\frac{1}{2} \cos ^{2}(2 t)+C\) (C) \(-\frac{1}{8} \sin \left(4 t^{2}\right)+C\) (D) \(\frac{1}{4} \sin \left(4 t^{2}\right)+C\)

Step-by-Step Solution

Answer

The answer is (A) \( \frac{1}{8} \sin(4t^2) + C \).

Step 1: Identify the integration technique

The given integral is \( \int t \cos(4t^2) \, dt \). Notice that the function inside the cosine is \( 4t^2 \), which suggests a substitution method, specifically a u-substitution, should be used to simplify the integral.

Step 2: Choose and apply a substitution

Let \( u = 4t^2 \). Then, the derivative is \( \frac{du}{dt} = 8t \), which implies \( dt = \frac{du}{8t} \). Substitute \( u \) and \( dt \) in the integral: \[ \int t \cos(4t^2) \, dt = \int t \cos(u) \cdot \frac{du}{8t} \]. Cancel the \( t \) and simplify to \[ \frac{1}{8} \int \cos(u) \cdot du \].

Step 3: Integrate with respect to u

The simplified integral \( \frac{1}{8} \int \cos(u) \, du \) can now be solved. The integral of \( \cos(u) \) is \( \sin(u) \), hence: \[ \frac{1}{8} \sin(u) + C \].

Step 4: Substitute back in terms of t

Since \( u = 4t^2 \), replace \( u \) with \( 4t^2 \) in the result: \[ \frac{1}{8} \sin(4t^2) + C \].

Step 5: Verify the solution against the options

The expression found \( \frac{1}{8} \sin(4t^2) + C \) matches option (A). Thus, the correct answer is (A).

Key Concepts

Integral Calculus

Integral calculus is a fundamental branch of mathematics that deals with integration, which is essentially the reverse process of differentiation. Integration is used to find quantities such as areas under curves, accumulated quantities, and other sums across an interval. When working with integrals, one can encounter:

• Definite integrals, which compute the net area within certain bounds.
• Indefinite integrals, which represent a family of functions and include a constant of integration, often denoted as C.

The integral in question is an indefinite integral, as evidenced by the presence of the constant C in the solution. Calculating an indefinite integral means finding an antiderivative of the function. In this case, we need a method, such as substitution, to manage the more complex expressions.

Cosine Function

The cosine function, denoted as \( \cos(x) \), is one of the primary trigonometric functions. It relates the adjacent side over the hypotenuse in a right triangle and has a number of important properties:

• The cosine function is periodic with a period of \( 2\pi \).
• Its values range from -1 to 1 for all real numbers \( x \).
• It is an even function, meaning \( \cos(-x) = \cos(x) \).

In the exercise, the function \( \cos(4t^2) \) reflects these properties but includes a more complicated argument \( 4t^2 \). This complication can make direct integration challenging, necessitating the substitution method. Here, \( \cos(u) \) is integrated easily compared to \( \cos(4t^2) \), emphasizing the utility of substitution in integration involving trigonometric functions.

Substitution Method

The substitution method, also known as \( u\)-substitution, is a powerful technique used in integral calculus to simplify the integration process. This method is particularly useful when the integral involves a composite function, where one function is nested inside another. Here's how it works:

• Identify a part of the integral that can be replaced by a single variable, often \( u \).
• Find the derivative of \( u \), expressed as \( du \), to replace \( dt \) or \( dx \).
• Substitute \( u \) and \( du \) into the integral to transform it into a simpler form.
• After integrating, revert back to the original variable by substituting the original expression of \( u \).

In this exercise, we set \( u = 4t^2 \). This substitution simplifies the integral to a familiar form, \( \frac{1}{8} \int \cos(u) \, du \), which can be directly integrated to \( \sin(u) \). By substituting back \( u = 4t^2 \), we obtain the solution in terms of \( t \). This method showcases its efficiency in tackling integrals with complex inner functions.