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

8. Definite Integrals

Substitution

Calculus

8. Definite Integrals

Substitution

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2:28 minutes

Problem 8.6.7

Textbook Question

7–84. Evaluate the following integrals.
7. ∫ from 0 to π/2 [sin θ / (1 + cos² θ)] dθ

Verified step by step guidance

Step 1: Recognize that the integral involves trigonometric functions. The integrand is sin(θ) / (1 + cos²(θ)). To simplify, consider substitution methods or trigonometric identities.

Step 2: Use the substitution u = cos(θ). Then, compute du = -sin(θ)dθ. This substitution will transform the integral into terms of u, making it easier to evaluate.

Step 3: Change the limits of integration according to the substitution. When θ = 0, u = cos(0) = 1. When θ = π/2, u = cos(π/2) = 0. The integral now has limits from u = 1 to u = 0.

Step 4: Rewrite the integral in terms of u. The original integral becomes ∫ from 1 to 0 [-du / (1 + u²)]. Notice the negative sign from du = -sin(θ)dθ.

Step 5: Reverse the limits of integration to remove the negative sign, resulting in ∫ from 0 to 1 [du / (1 + u²)]. This integral is now in a standard form, which can be evaluated using the arctangent function: ∫ [du / (1 + u²)] = arctan(u).

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Key Concepts

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

Definite Integrals

A definite integral calculates the accumulation of a quantity, represented as the area under a curve, between two specified limits. In this case, the integral is evaluated from 0 to π/2, meaning we are interested in the area under the curve of the function sin(θ) / (1 + cos²(θ)) from θ = 0 to θ = π/2.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, relate angles to ratios of sides in right triangles. Understanding these functions is crucial for evaluating integrals involving them, as they often appear in calculus problems. The integral in question involves sin(θ) and cos²(θ), which are fundamental in trigonometric identities and integration techniques.

Integration Techniques

Integration techniques are methods used to find the integral of a function, which can include substitution, integration by parts, or recognizing standard forms. For the given integral, recognizing the structure of the integrand and possibly using a substitution can simplify the evaluation process, making it easier to compute the area under the curve.