Answer:
[tex]\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, dx = -ln(2)[/tex]
General Formulas and Concepts:
Calculus
Differentiation
- Derivatives
- Derivative Notation
Trig Derivatives
Integration
- Integrals
- Definite 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
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, 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 = -sin(x) \ dx[/tex]
- [Bounds of Integration] Change: [tex]\displaystyle [\frac{1}{2}, 1][/tex]
Step 3: Integrate Pt. 2
- [Integral] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, dx = -\int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{-sin \ x}{1 + cos \ x}} \, dx[/tex]
- [Integral] U-Substitution: [tex]\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, dx = -\int\limits^{1}_{\frac{1}{2}} {\frac{1}{u}} \, du[/tex]
- [Integral] Logarithmic Integration: [tex]\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, dx = -(ln|u|) \bigg| \limits^{1}_{\frac{1}{2}}[/tex]
- Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, dx = -ln(2)[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
Book: College Calculus 10e
