Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

7. Antiderivatives & Indefinite Integrals

Integrals of Trig Functions

Calculus

7. Antiderivatives & Indefinite Integrals

Integrals of Trig Functions

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1:10 minutes

Problem 8.1.68

Textbook Question

68. Different methods
a. Evaluate ∫(cot x csc² x) dx using the substitution u=cotx.

Verified step by step guidance

Step 1: Identify the substitution. Let u = cot(x). This substitution simplifies the integral by replacing cot(x) with u.

Step 2: Compute the derivative of u with respect to x. Since u = cot(x), we know that du/dx = -csc²(x). Rearrange this to express dx in terms of du: dx = -du/csc²(x).

Step 3: Rewrite the integral in terms of u. Substitute u = cot(x) and dx = -du/csc²(x) into the integral ∫(cot(x) csc²(x)) dx. This gives ∫(u csc²(x) * (-du/csc²(x))).

Step 4: Simplify the integral. Notice that csc²(x) cancels out, leaving ∫(-u du). This simplifies the integral to -∫u du.

Step 5: Integrate with respect to u. The integral of u with respect to u is (u²/2), so the result becomes -(u²/2). Finally, substitute back u = cot(x) to express the solution in terms of x: -(cot²(x)/2).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Integration

Integration is a fundamental concept in calculus that involves finding the integral of a function, which represents the area under the curve of that function. It can be thought of as the reverse process of differentiation. In this context, we are tasked with evaluating an integral, which requires understanding how to manipulate and simplify expressions to find their antiderivatives.

Substitution Method

The substitution method is a technique used in integration to simplify the process by changing variables. By substituting a new variable (in this case, u = cot x), we can transform the integral into a more manageable form. This method is particularly useful when dealing with composite functions, allowing us to express the integral in terms of a single variable.

Trigonometric Functions

Trigonometric functions, such as cotangent (cot) and cosecant (csc), are essential in calculus, especially in integrals involving angles. Understanding their properties and relationships is crucial for manipulating expressions involving these functions. In this problem, recognizing how cot x and csc² x relate to each other will aid in simplifying the integral after substitution.