Problem 44 - Thomas Calculus

Question

In Exercises \(17-54\) , find the most general antiderivative or indefinite integral. Check your answers by differentiation. $$ \int \frac{1}{2}\left(\csc ^{2} x-\csc x \cot x\right) d x $$

Step-by-Step Solution

Answer

The most general antiderivative is \(-\frac{1}{2} \cot x + \frac{1}{2} \csc x + C.\)

Step 1: Integrate the Expression

The problem is to find the indefinite integral of \( \frac{1}{2}(\csc^2 x - \csc x \cot x) \) with respect to \( x \). We break the integral into two parts based on the properties of definite integrals:\[\int \frac{1}{2}(\csc^2 x - \csc x \cot x) \, dx = \frac{1}{2} \int \csc^2 x \, dx - \frac{1}{2} \int \csc x \cot x \, dx.\]We know: \( \int \csc^2 x \, dx = -\cot x + C \) and \( \int \csc x \cot x \, dx = -\csc x + C \).Then:\[\frac{1}{2} \int \csc^2 x \, dx = -\frac{1}{2} \cot x + C,\]\[\frac{1}{2} \int \csc x \cot x \, dx = -\frac{1}{2} \csc x + C.\]

Step 2: Combine the Integrals

Now, we sum the results of the two individual integrals obtained in Step 1:\[-\frac{1}{2} \cot x + C - \left( -\frac{1}{2} \csc x + C \right) = -\frac{1}{2} \cot x + \frac{1}{2} \csc x + C.\]This is the most general form of the antiderivative for the given integral.

Step 3: Differentiate to Verify

To ensure the correctness of the found antiderivative, we differentiate it:- Differentiate \(-\frac{1}{2} \cot x + \frac{1}{2} \csc x\) with respect to \( x \).- The derivative of \(-\frac{1}{2} \cot x\) is \( \frac{1}{2} \csc^2 x \), and the derivative of \(\frac{1}{2} \csc x \) is \(-\frac{1}{2} \csc x \cot x \).- Combine these results:\[\frac{1}{2} \csc^2 x - \frac{1}{2} \csc x \cot x,\]which matches the original integrand.

Key Concepts

Indefinite Integrals

Indefinite integrals are like reverse operations to differentiation, often referred to as antiderivatives. When we compute the indefinite integral of a function, we are essentially determining the family of functions whose derivative gives back the original function. An indefinite integral is symbolized by the integral sign (\(\int\)), followed by the function to be integrated, and the differential of the variable. It's written as:

• \[\int f(x) \, dx\] represents the antiderivative of function \(f(x)\).

Remember, indefinite integrals include a constant of integration, denoted by \(C\), because differentiation makes the constant disappear. This constant ensures that every possible antiderivative is captured. In our example, solving \(\int \frac{1}{2}(\csc^2 x - \csc x \cot x) \, dx\) involves integrating each term separately to get the most general antiderivative representation.

Trigonometric Integrals

Trigonometric integrals often involve integrating functions composed of trigonometric operations like sine, cosine, tangent, cotangent, secant, and cosecant. These integrals may appear intimidating, but with knowledge of certain antiderivatives, they become manageable. Among these, there are some standard integrals:

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

To tackle the given integral \(\frac{1}{2}(\csc^2 x - \csc x \cot x)\), we split it into its components and utilize known trigonometric antiderivatives to integrate each term. Applying these known results made the integration process straightforward, highlighting the usefulness of memorizing fundamental trigonometric integrals.

Antiderivatives

Antiderivatives are the backbone of calculus integration. They are the reverse of derivatives, offering a way to construct helper functions that tell us about the area under a curve. All indefinite integrals are antiderivatives.When calculating an antiderivative, the constant \(C\) plays a crucial role:

• It represents an entire family of curves.
• Shows that infinitely many functions have the same derivative.

Our specific case involved calculating the antiderivative of \(\frac{1}{2}(\csc^2 x - \csc x \cot x)\). By finding the antiderivative, we determined that the most general function family satisfying the equation can be expressed as: \[-\frac{1}{2} \cot x + \frac{1}{2} \csc x + C\] This result was verified by differentiating it back, which brought us to the original integrand. Such verification helps solidify understanding and correctness of antiderivative solutions.