Problem 25 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the indefinite integral. \(\int \sqrt{\cot x} \csc ^{2} x d x\)

Step-by-Step Solution

Answer

The integral evaluates to \(-\frac{2}{3} \csc^{3/2} x + C\).

Step 1: Simplify the Integrand

Firstly, identify the parts within the integral that can be simplified. Here, you’ve got \( \sqrt{\cot x} \cdot \csc ^{2} x \). Since \( \csc x = \frac{1}{\sin x} \), and \( \cot x = \frac{\cos x}{\sin x} \), we have \( \sqrt{\cot x} = \sqrt{\frac{\cos x}{\sin x}} = \frac{\sqrt{\cos x}}{\sqrt{\sin x}} \). And the \( \csc^2 x = \frac{1}{\sin^2 x} \), multiplying with \( \sqrt{\cot x} \), we get the integrand to simplify as \( \frac{\sqrt{\cos x}}{\sin^{5/2} x} \).

Step 2: Use Substitution Method

Let's apply substitution. Let \( u = \sin x \). Then \( du = \cos x \, dx \). This makes \( \sqrt{\cos x} \, dx = \frac{1}{\sqrt{\sin x}} \, du \). Substitute back into the integrand to get \( \int u^{-5/2} \, du \).

Step 3: Integrate with respect to u

Now, integrate \( \int u^{-5/2} \, du \). Use the power rule for integration which states that \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \). Here, \( n = -\frac{5}{2} \), so:\[\int u^{-5/2} \, du = \frac{u^{-5/2 + 1}}{-5/2 + 1} + C = \frac{u^{-3/2}}{-3/2} + C = -\frac{2}{3}u^{-3/2} + C\]

Step 4: Substitute Back to x

Replace \( u \) back with \( \sin x \), which was our substitution. Thus:\[-\frac{2}{3}(\sin x)^{-3/2} + C\]This is equivalent to:\[-\frac{2}{3}\cdot\frac{1}{(\sin x)^{3/2}} + C\]or:\[-\frac{2}{3} \csc^{3/2} x + C \]

Step 5: Final Answer

The solution to the indefinite integral \( \int \sqrt{\cot x} \csc ^{2} x \, dx \) is:\[-\frac{2}{3} \csc^{3/2} x + C\]where \( C \) is the constant of integration.

Key Concepts

Indefinite Integral

An indefinite integral represents a family of functions and is the reverse process of differentiation. Unlike definite integrals, indefinite integrals do not evaluate to a number; instead, they provide a general form of antiderivatives with an added constant of integration. This constant, denoted by C, accounts for the fact that there are potentially infinite antiderivatives for a given function.

When evaluating indefinite integrals, the aim is to find a function F(x) whose derivative matches the integrand, denoted as \( \int f(x) \, dx = F(x) + C \). Understanding the properties of indefinite integrals is crucial for solving various calculus problems and provides a gateway to more complex integration techniques.

Trigonometric Substitution

Trigonometric substitution is a technique used to simplify integrals, especially those involving square roots or trigonometric identities. This method involves replacing variables with trigonometric identities, making the algebra more manageable.

In our exercise, we used the substitution \( u = \sin x \), allowing us to express the integrand in terms of u and simplify the integral substantially. The substitution transforms \( \cos x \, dx \) into \( \frac{1}{\sqrt{\sin x}} \, du \), helping to reduce the complexity of the integral. The ultimate goal with substitution is to create an integrable function that can be directly solved using basic integration rules.

Power Rule for Integration

The power rule for integration is a fundamental technique that helps integrate functions of the form \( u^n \). The rule is simple yet powerful:

• For any real number \( n eq -1 \), the integral of \( u^n \) is: \[ \int u^n \, du = \frac{u^{n+1}}{n+1} + C \]

This rule is extensively used due to its straightforward application on polynomial and simple rational functions.

In the exercise, applying the power rule converted the expression \( \int u^{-5/2} \, du \) to \( -\frac{2}{3}u^{-3/2} + C \). Understanding and mastering this rule can significantly enhance one's ability to tackle more complex integrals.

Trigonometric Functions

Trigonometric functions such as sine and cosine are fundamental in calculus, offering a way to handle curves and oscillations within integrals. These functions come with specific identities and properties, essential for solving trigonometric integrals.

In our problem, simplifying \( \sqrt{\cot x} \cdot \csc ^{2} x \) into known trigonometric terms allowed for efficient integration. Recognizing \( \csc x \) and \( \cot x \) in their fundamental forms, \( \csc x = \frac{1}{\sin x} \) and \( \cot x = \frac{\cos x}{\sin x} \), sets the stage for using these identities in integration techniques. Mastery of trigonometric identities is crucial as they frequently appear in various integration problems, providing the groundwork for successful solution strategies.