Respuesta :

Space

Answer:

[tex]\displaystyle \int\limits^{\frac{\pi}{3}}_0 {\cos^2 \theta \sin \theta} \, d\theta = \boxed{ \frac{7}{24} }[/tex]

General Formulas and Concepts:
Calculus

Differentiation

  • Derivatives
  • Derivative Notation

Integration

  • Integrals

Integration Rule [Reverse Power Rule]:
[tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]

Integration Rule [Fundamental Theorem of Calculus 1]:
[tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]

Integration Methods: U-Substitution and U-Solve

Step-by-step explanation:

Step 1: Define

Identify given.

[tex]\displaystyle \int\limits^{\frac{\pi}{3}}_0 {\cos^2 \theta \sin \theta} \, d\theta[/tex]

Step 2: Integrate Pt. 1

Identify variables for u-substitution/u-solve.

  1. Set u:
    [tex]\displaystyle u = \cos \theta[/tex]
  2. [u] Differentiate [Trigonometric Differentiation]:
    [tex]\displaystyle du = - \sin \theta \ d\theta[/tex]
  3. [du] Rewrite [U-Solve]
    [tex]\displaystyle d\theta = - \csc \theta \ du[/tex]
  4. [Limits] Swap:
    [tex]\displaystyle \left \{ {{ \theta = \frac{\pi}{3} } \atop { \theta = 0 }} \right. \rightarrow \left \{ {{ u = \cos \bigg( \frac{\pi}{3} \bigg) } \atop { u = \cos 0}} \right. \rightarrow \left \{ {{ u = \frac{1}{2} } \atop { u = 1 }} \right.[/tex]

Step 3: Integrate Pt. 2

  1. [Integrand] Rewrite:
    [tex]\displaystyle \begin{aligned}\int\limits^{\frac{\pi}{3}}_0 {\cos^2 \theta \sin \theta} \, d\theta & = \int\limits^{\frac{\pi}{3}}_0 {\cos \theta \cos \theta \sin \theta} \, d\theta \\\end{aligned}[/tex]
  2. [Integral] Apply Integration Method [U-Solve]:
    [tex]\displaystyle \begin{aligned}\int\limits^{\frac{\pi}{3}}_0 {\cos^2 \theta \sin \theta} \, d\theta & = \int\limits^{\frac{\pi}{3}}_0 {\cos \theta \cos \theta \sin \theta} \, d\theta \\& = \int\limits^{\frac{1}{2}}_0 {u^2 \sin \theta \csc \theta} \, du \\\end{aligned}[/tex]
  3. [Integrand] Simplify:
    [tex]\displaystyle \begin{aligned}\int\limits^{\frac{\pi}{3}}_0 {\cos^2 \theta \sin \theta} \, d\theta & = \int\limits^{\frac{\pi}{3}}_0 {\cos \theta \cos \theta \sin \theta} \, d\theta \\& = \int\limits^{\frac{1}{2}}_0 {u^2 \sin \theta \csc \theta} \, du \\& = \int\limits^{\frac{1}{2}}_0 {u^2} \, du \\\end{aligned}[/tex]
  4. [Integral] Apply Integration Rule [Reverse Power Rule]:
    [tex]\displaystyle \begin{aligned}\int\limits^{\frac{\pi}{3}}_0 {\cos^2 \theta \sin \theta} \, d\theta & = \int\limits^{\frac{\pi}{3}}_0 {\cos \theta \cos \theta \sin \theta} \, d\theta \\& = \int\limits^{\frac{1}{2}}_0 {u^2 \sin \theta \csc \theta} \, du \\& = \int\limits^{\frac{1}{2}}_0 {u^2} \, du \\& = \frac{u^3}{3} \bigg| \limits^{\frac{1}{2}}_0 \\\end{aligned}[/tex]
  5. Apply Integration Rule [Fundamental Theorem of Calculus 1]:
    [tex]\displaystyle \begin{aligned}\int\limits^{\frac{\pi}{3}}_0 {\cos^2 \theta \sin \theta} \, d\theta & = \int\limits^{\frac{\pi}{3}}_0 {\cos \theta \cos \theta \sin \theta} \, d\theta \\& = \int\limits^{\frac{1}{2}}_0 {u^2 \sin \theta \csc \theta} \, du \\& = \int\limits^{\frac{1}{2}}_0 {u^2} \, du \\& = \frac{u^3}{3} \bigg| \limits^{\frac{1}{2}}_0 \\& = \frac{\bigg( \frac{1}{2} \bigg) ^3}{3} - \frac{0}{3} \\\end{aligned}[/tex]
  6. Evaluate:
    [tex]\displaystyle \begin{aligned}\int\limits^{\frac{\pi}{3}}_0 {\cos^2 \theta \sin \theta} \, d\theta & = \int\limits^{\frac{\pi}{3}}_0 {\cos \theta \cos \theta \sin \theta} \, d\theta \\& = \int\limits^{\frac{1}{2}}_0 {u^2 \sin \theta \csc \theta} \, du \\& = \int\limits^{\frac{1}{2}}_0 {u^2} \, du \\& = \frac{u^3}{3} \bigg| \limits^{\frac{1}{2}}_0 \\& = \frac{\bigg( \frac{1}{2} \bigg) ^3}{3} - \frac{0}{3} \\& = \boxed{ \frac{7}{24} }\end{aligned}[/tex]

∴ the exact value of the definite integral is equal to 7/24.

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Learn more about integration: https://brainly.com/question/24705479

Learn more about Calculus: https://brainly.com/question/27593180

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Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

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