Answer:
[tex]\displaystyle \int\limits^{\frac{2 \pi}{3}}_{\frac{- \pi}{3}} {\frac{sinx}{1 + cosx}} \, dx = ln3[/tex]
General Formulas and Concepts:
Pre-Calculus
Calculus
Derivatives
Derivative Notation
Trig Derivatives
Integrals
Integration Constant C
Integration Rule [Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]
Integration Property [Multiplied Constant]: [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]
Trig Integration
Logarithmic Integration
U-Substitution
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle \int\limits^{\frac{2 \pi}{3}}_{\frac{- \pi}{3}} {\frac{sinx}{1 + cosx}} \, dx[/tex]
Step 2: Integrate Pt. 1
Identify variables for u-substitution.
- Set u: [tex]\displaystyle u = 1 + cos(x)[/tex]
- [u] Differentiate [Trig Derivative]: [tex]\displaystyle du = -sinx \ dx[/tex]
Step 3: Integrate Pt. 2
- [Integral] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle \int\limits^{\frac{2 \pi}{3}}_{\frac{- \pi}{3}} {\frac{sinx}{1 + cosx}} \, dx = -\int\limits^{\frac{2 \pi}{3}}_{\frac{- \pi}{3}} {\frac{-sinx}{1 + cosx}} \, dx[/tex]
- [Integral] U-Substitution: [tex]\displaystyle \int\limits^{\frac{2 \pi}{3}}_{\frac{- \pi}{3}} {\frac{sinx}{1 + cosx}} \, dx = -\int\limits^{\frac{1}{2}}_{\frac{3}{2}} {\frac{1}{u}} \, du[/tex]
- [Integral] Logarithmic Integration: [tex]\displaystyle \int\limits^{\frac{2 \pi}{3}}_{\frac{- \pi}{3}} {\frac{sinx}{1 + cosx}} \, dx = -ln|u| \bigg| \limits^{\frac{1}{2}}_{\frac{3}{2}}[/tex]
- Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^{\frac{2 \pi}{3}}_{\frac{- \pi}{3}} {\frac{sinx}{1 + cosx}} \, dx = -(-ln3)[/tex]
- Simplify: [tex]\displaystyle \int\limits^{\frac{2 \pi}{3}}_{\frac{- \pi}{3}} {\frac{sinx}{1 + cosx}} \, dx = ln3[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
Book: College Calculus 10e
