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Chapter 5: Problem 28

Evaluate the integral and check your answer by differentiating. $$\int \frac{d y}{\csc y}$$

Short Answer

Expert verified
The integral is \(-\cos y + C\). Verified by differentiation.

Step by step solution

01

Rewrite the Integral

The integrand is \( \frac{1}{\csc y} \). Recall that \( \csc y = \frac{1}{\sin y} \). Therefore, \( \frac{1}{\csc y} = \sin y \). So, the integral becomes \( \int \sin y \, dy \).
02

Integrate \( \sin y \)

The integral of \( \sin y \) with respect to \( y \) is \(-\cos y \). Therefore, \( \int \sin y \, dy = -\cos y + C \), where \( C \) is the constant of integration.
03

Check by Differentiating

To verify our result, differentiate \(-\cos y + C\). The derivative of \(-\cos y\) is \( -(-\sin y) = \sin y \), and the derivative of the constant \( C \) is 0. This gives \( \sin y \), which matches the original integrand \( \frac{1}{\csc y} = \sin y \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration Techniques
Integration techniques are methods used to evaluate integrals, and they are fundamental in calculus. In this exercise, we applied the substitution by recognizing a reciprocal identity.
Recognizing trigonometric identities can simplify the integration process. For instance, recognizing that \( \csc y \) is the same as \( \frac{1}{\sin y} \) allows us to rewrite the function inside the integral for easier solving.
Once reformulated, you can perform the integration using basic rules. Knowing the integral of \( \sin y \) simplifies our task greatly. By knowing these identities and techniques, complex integrals become manageable.
  • Use trigonometric identities to transform integrals.
  • Apply basic integration rules for common functions.
  • Don't forget the constant of integration \( C \).
Trigonometric Functions
Trigonometric functions like \( \csc y \), \( \sin y \), and \( \cos y \) are core components of both integration and differentiation in calculus. Understanding these functions is crucial.
In this example, the function \( \csc y \) was converted to \( \sin y \) because it simplified the integration process. The function \( \sin y \) has a straightforward antiderivative, which is \(-\cos y\).
  • \( \csc y = \frac{1}{\sin y} \) for simplification.
  • Integrate \( \sin y \) to find \(-\cos y\).
  • Trigonometric functions often recur in calculus problems, affecting how you solve them.
Familiarity with these conversions allows for seamless transitions between forms during calculations.
Differentiation
Differentiation is the reverse process of integration and verifies our integrated results. It involves finding the derivative of a function with respect to a variable.
To check the correctness of an integral, differentiate it. For this exercise, the expression \(-\cos y + C\) was differentiated.
Differentiating \(-\cos y\) gave \( \sin y \), matching the original integrand \( \sin y \). This confirms that the integration was performed correctly.
  • Verify integrals through differentiation.
  • Derivatives of constants are 0.
  • Confirm the process with inverse differentiation.
Differentiation provides a practical method to ensure calculated integrals are accurate.

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Most popular questions from this chapter

Use a calculating utility to find the midpoint approximation of the integral using \(n=20\) sub-intervals, and then find the exact value of the integral using Part 1 of the Fundamental Theorem of Calculus. $$\int_{0}^{\pi / 2} \sin x d x$$

Use Part 2 of the Fundamental Theorem of Calculus to find the derivatives. (a) \(\frac{d}{d x} \int_{0}^{x} \frac{d t}{1+\sqrt{t}}\) (b) \(\frac{d}{d x} \int_{1}^{x} \ln t d t\)

Solve the initial-value problems. $$\frac{d y}{d x}=\sqrt{5 x+1}, y(3)=-2$$

Determine whether the statement is true or false. Explain your answer. If \(f(x)\) is continuous on the interval \([a, b],\) and if the definite integral of \(f(x)\) over this interval has value 0 , then the equation \(f(x)=0\) has at least one solution in the interval \([a, b]\)

Evaluate the integrals using appropriate substitutions. $$\int \frac{y}{\sqrt{2 y+1}} d y$$
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