Problem 21 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the indefinite integral. \(\int \frac{\cos x}{\sin ^{2} x} d x\)

Step-by-Step Solution

Answer

-\csc x + C

Step 1: Understanding the Integral

The integral given is \( \int \frac{\cos x}{\sin^2 x} \, dx \). We need to find an antiderivative for the function \( \frac{\cos x}{\sin^2 x} \). Notice that \( \frac{\cos x}{\sin^2 x} \) can be rewritten as \( \cos x \cdot \csc^2 x \).

Step 2: Choosing a Substitution

To simplify the integral, let's use substitution. Assume \( u = \sin x \). Then, the derivative \( du = \cos x \, dx \). This substitution will simplify the integral significantly.

Step 3: Performing the Substitution

Rewrite the integral \( \int \frac{\cos x}{\sin^2 x} \, dx \) using the substitution. It becomes \( \int \frac{1}{u^2} \, du \), as \( \cos x \, dx = du \) and \( \sin x = u \).

Step 4: Integrating the Simplified Expression

Now, integrate \( \int \frac{1}{u^2} \, du \), which is \( \int u^{-2} \, du \). The antiderivative of \( u^{-2} \) is \( -u^{-1} + C \), where \( C \) is the constant of integration.

Step 5: Substituting Back

Substitute back \( u = \sin x \) into the expression \( -u^{-1} + C \). This results in \( -\frac{1}{\sin x} + C \). Since \( \csc x = \frac{1}{\sin x} \), the integral simplifies to \( -\csc x + C \).

Key Concepts

Substitution Method

The Substitution Method is a powerful technique for evaluating integrals, especially when dealing with more complex functions. It involves changing variables to simplify the integral into a more manageable form. In our original exercise, we start by setting a new variable, say \( u \), to replace a part of the original integrand. Consider our integral \( \int \frac{\cos x}{\sin^2 x} \, dx \). By noticing that \( u = \sin x \), we link it directly to the integration's structure. This choice is not random; we make it because the derivative of \( \sin x \) is \( \cos x \), which also appears in the numerator.

• After setting \( u = \sin x \), we find that \( du = \cos x \, dx \). This allows us to transform the integral entirely in terms of \( u \).
• The integral then becomes \( \int \frac{1}{u^2} \, du \), greatly simplifying our task.

The effectiveness of substitution hinges on recognizing these patterns and simplifying the integral into something much easier to compute.

Trigonometric Integrals

Trigonometric Integrals arise frequently in calculus, often involving functions such as sine, cosine, tangent, and their reciprocals. They can appear complex, but there are systematic methods to solve them. In our exercise, we deal with a trigonometric function \( \frac{\cos x}{\sin^2 x} \). To handle such a form, look for relationships between the trigonometric functions involved:

• First, note \( \sin x \) and \( \cos x \), and how they interact with their reciprocal identities. Here, \( \csc x = \frac{1}{\sin x} \), thus \( \csc^2 x = \frac{1}{\sin^2 x} \).
• This recognition transforms the integrand into \( \cos x \cdot \csc^2 x \), which the substitution method then further simplifies.

Understanding these relationships is essential when tackling trigonometric integrals, allowing us to rewrite and solve them efficiently.

Antiderivatives

Antiderivatives are the reverse process of derivatives, representing the family of functions that have a given function as their derivative. In indefinite integrals, finding the antiderivative is key. Think of it as unwinding the derivative process. In our equation, once simplified through substitution, we need to integrate \( \int u^{-2} \, du \).

• The antiderivative of \( u^{-n} \) where \( n eq 1 \) is \( -\frac{1}{u} + C \). In this case, it results in \( -u^{-1} + C \), where \( C \) is the integration constant.
• Once integrated, substituting back \( u = \sin x \) gives the initial function's antiderivative, simplifying to \( -\frac{1}{\sin x} + C \), or \( -\csc x + C \).

Understanding antiderivatives allows us to find integral solutions and complete the original calculus operation.