Problem 39 - Calculus Single Variable

Question

Use trigonometric identities to compute the indefinite integrals. $$ \int \cos ^{2}(x) d x $$

Step-by-Step Solution

Answer

\( \int \cos^2(x) \, dx = \frac{1}{2}x + \frac{1}{4}\sin(2x) + C \).

Step 1: Identify the Trigonometric Identity

To compute the integral \( \int \cos^2(x) \, dx \), we first need to recall a trigonometric identity that can simplify \( \cos^2(x) \). The identity we use is the power-reduction formula for cosine: \( \cos^2(x) = \frac{1 + \cos(2x)}{2} \).

Step 2: Substitute the Identity into the Integral

Replace \( \cos^2(x) \) in the integral with \( \frac{1 + \cos(2x)}{2} \). This gives us the integral: \( \int \frac{1 + \cos(2x)}{2} \, dx \).

Step 3: Split the Integral

Split the integral \( \int \frac{1 + \cos(2x)}{2} \, dx \) into two separate integrals: \( \int \frac{1}{2} \, dx + \int \frac{\cos(2x)}{2} \, dx \). This simplifies the computation process.

Step 4: Integrate Each Part Separately

Compute each integral separately:1. For \( \int \frac{1}{2} \, dx \), the result is \( \frac{1}{2}x \).2. For \( \int \frac{\cos(2x)}{2} \, dx \), apply a substitution \( u = 2x \) making \( du = 2 \, dx \), or equivalently \( dx = \frac{1}{2} du \). The integral becomes \( \frac{1}{4} \int \cos(u) \, du \). The result of \( \int \cos(u) \, du \) is \( \sin(u) \), so we have \( \frac{1}{4} \sin(2x) \).

Step 5: Combine the Results

Combine the results of the separate integrals: \( \int \cos^2(x) \, dx = \frac{1}{2}x + \frac{1}{4}\sin(2x) + C \), where \( C \) is the constant of integration.

Key Concepts

Trigonometric Identities

Trigonometric identities are mathematical equations that express relationships between trigonometric functions. They are incredibly useful for simplifying complex trigonometric expressions and are key to solving many calculus problems.
One crucial identity used in integrating powers of trigonometric functions is the identity for cosine squared:

• \( \cos^2(x) = \frac{1 + \cos(2x)}{2} \)

This identity is called a power-reduction formula and is special because it allows us to reduce the power of cosine, simplifying integration. Remembering and understanding these identities can make solving calculus problems more manageable.

Power-Reduction Formula

The power-reduction formula is a tool to simplify expressions involving trigonometric functions raised to an exponent. It specifically helps us to express these functions using identities that are easier to integrate. For the cosine function, the power-reduction formula is:

• The formula: \( \cos^2(x) = \frac{1 + \cos(2x)}{2} \)

Using this formula, expressions with squared trigonometric functions can be reduced to a combination of constant and doubled angle trigonometric functions. This simplifies the integration process by converting a potentially complicated function into a simpler form. As seen in the step-by-step solution, applying this transformation allows us to divide the original integral into separate, simpler parts. Utilizing power-reduction formulas effectively is a cornerstone skill in calculus.

U-Substitution

U-substitution is a technique used in calculus to simplify the process of integration. Particularly useful in the context of indefinite integrals, it involves substituting part of the integral variable with a new variable, often simplifying the integral significantly. Here's how it can be applied:

• First, identify a substitution: in our problem, we used \( u = 2x \).
• Compute the derivative: this gives \( du = 2 \, dx \).
• Rewrite \( dx \) in terms of \( du \), which gives \( dx = \frac{1}{2} du \).
• Substitute into the integral: \( \int \frac{\cos(2x)}{2} \, dx \) becomes \( \frac{1}{4} \int \cos(u) \, du \).

This method turns an integral that might seem complex into a more straightforward integral that can be directly computed. Mastering u-substitution empowers you to tackle a wider range of integral problems with increased confidence.

Calculus

Calculus is the branch of mathematics that deals with rates of change (differential calculus) and accumulation (integral calculus). It is a critical tool in science, engineering, and many fields where dynamic systems are studied.
Indefinite integrals, a focus of this topic, can be thought of as the reverse operation of differentiation. This means we are looking for functions whose derivatives provide the function within the integral. For example:

• The integral \( \int \cos(x) \, dx = \sin(x) + C \)
• This antiderivative expresses how the rate of change for the sine function corresponds to the cosine function we started with.

In the problem of finding \( \int \cos^2(x) \, dx \), calculus allows us to determine that \( \frac{1}{2}x + \frac{1}{4}\sin(2x) + C \) is the antiderivative, meaning its derivative returns the original function \( \cos^2(x) \). This illustrates how calculus concepts underpin practical problem-solving in mathematics.