Problem 45 - Thomas Calculus

Question

In Exercises \(17-54\) , find the most general antiderivative or indefinite integral. Check your answers by differentiation. $$ \int\left(\sin 2 x-\csc ^{2} x\right) d x $$

Step-by-Step Solution

Answer

The antiderivative is \( -(1/2) \cos 2x - \cot x + C \).

Step 1: Identify Individual Integral Components

The given integral \( \int(\sin 2x - \csc^2 x) \, dx \) consists of two parts: \( \int \sin 2x \, dx \) and \( \int - \csc^2 x \, dx \). We will find the antiderivative of each component separately.

Step 2: Calculate Antiderivative of First Integral Component

To find \( \int \sin 2x \, dx \), we use the substitution method. Let \( u = 2x \), then \( du = 2 \, dx \) or \( (1/2)du = dx \). The integral becomes \( \int (1/2) \sin u \, du = -(1/2) \cos u + C_1 \). Reverting \( u \) back to \( 2x \), we get \( -(1/2) \cos 2x + C_1 \).

Step 3: Calculate Antiderivative of Second Integral Component

The integral \( \int - \csc^2 x \, dx \) is a standard integral with the antiderivative \( \cot x + C_2 \) (the negative sign remains outside). Therefore, it evaluates to \( -\cot x + C_2 \).

Step 4: Combine Antiderivatives into Single Expression

The antiderivatives of the two components are combined: \( -(1/2) \cos 2x - \cot x + C \), where \( C = C_1 + C_2 \) is a constant of integration.

Step 5: Verify by Differentiation

Differentiate the result \( -(1/2) \cos 2x - \cot x + C \) with respect to \( x \). The derivative is \((d/dx) [-(1/2) \cos 2x] = \sin 2x\) and \((d/dx)[-\cot x] = \csc^2 x\). Hence, the differentiation gives \( \sin 2x - \csc^2 x \), confirming our original integrand.

Key Concepts

Understanding Antiderivatives

In calculus, finding the antiderivative or indefinite integral of a function means discovering another function whose derivative is the original function you started with. It is like reversing the process of differentiation. When you integrate a function, you are looking for all such functions that, when differentiated, give back the function you are working with.
\[\int f(x) \, dx = F(x) + C\]
Here, \(F(x)\) is the antiderivative of \(f(x)\), and \(C\) is the constant of integration which accounts for the family of functions differing by a constant value. It's crucial to include this constant because integration can lead to multiple functions, each varying by a constant.
A major goal in integration is to find the most general antiderivative, ensuring you've identified all functions that could produce the original function when derivated. This principle is the foundational step before dealing with more complex integration techniques.

Substitution Method Simplified

The substitution method is a powerful technique used to simplify integrals that appear too complex to integrate directly. It involves substituting part of the integral with a new variable, often making the integral more manageable.

• Identify a part of the integral to substitute. This is typically an expression whose derivative is present as another factor in the integral.
• Substitute with a new variable like \(u\), and find \(du\), the differential of \(u\). Adjust the integral accordingly.
• Perform the integral in terms of this new variable.
• Finally, revert the substitution by replacing \(u\) back to the original variable.

For example, in our problem, substituting \(u = 2x\) simplifies \( \int \sin 2x \, dx\) to \( -(1/2) \cos u + C_1\), which is significantly easier to integrate. This method aligns with reversing differentiation rules and can be a simple yet powerful tool to tackle integrals.

Exploring Trigonometric Integrals

Trigonometric Integrals are a common category of integrals involving trigonometric functions, such as sine, cosine, and others. Integrating these can often entail using trigonometric identities and rules to simplify the function first.
For example, common integrals include:

• \(\int \sin x \, dx = -\cos x + C\)
• \(\int \cos x \, dx = \sin x + C\)
• \(\int \csc^2 x \, dx = -\cot x + C\)

In the exercise, the integral of \(-\csc^2 x\) directly gives \(-\cot x + C\), using the well-known derivative relationship of \(\csc^2 x\) to keep integration straightforward. Understanding these relationships and identities allows for efficient computation and manipulation of more complex integrals.

Mastering Integration by Parts

Integration by Parts is a technique based on the product rule for differentiation and is particularly useful for integrals involving products of different types of functions. The formula is:
\[\int u \, dv = uv - \int v \, du\]
Here's how you apply it:

• Identify parts of the integral matching \(u\) and \(dv\).
• Differentiate \(u\) to find \(du\), and integrate \(dv\) to find \(v\).
• Plug into the integration by parts equation to simplify.

Although not used directly in the given problem, understanding integration by parts is invaluable as it provides a systematic way to break down and solve more complex integrals that are resistant to direct integration or basic substitution, such as products of polynomials and exponential or trigonometric functions.