Answer:
[tex]\int\limits^1 _\frac{1}{2} {\frac{-1}{u} } \, du[/tex]
General Formulas and Concepts:
Pre-Algebra
- Order of Operations: BPEMDAS
Calculus
- Derivative of trig: [tex]\frac{d}{dx} [cos(x)] = -sin(x)[/tex]
- Integral of 1/u: [tex]\int\limits {\frac{1}{u} } \, du = ln|u|[/tex]
- Fundamental Theory of Calculus Part 1: [tex]\int\limits^a_b {f(x)} \, dx = F(b)-F(a)[/tex]
- Integral Solving Methods - u-substitution
Step-by-step explanation:
Step 1: Define
[tex]\int\limits^\frac{-\pi}{2} _\frac{-2\pi}{3} {\frac{sin(x)}{1+cos(x)} } \, dx[/tex]
Step 2: Evaluate
- Define a u: u = 1 + cos(x)
- Define du: du = -sin(x)dx
- Rewrite integral: [tex]-\int\limits^\frac{-\pi}{2} _\frac{-2\pi}{3} {\frac{-sin(x)}{1+cos(x)} } \, dx[/tex]
- u-substitute: [tex]-\int\limits^\frac{-\pi}{2} _\frac{-2\pi}{3} {\frac{1}{u} } \, du[/tex]
- Change limits [a]: a = 1 + cos(-π/2) = 1
- Change limits [b]: b = 1 + cos(-2π/3) = 1/2
- Change limits: [tex]-\int\limits^1 _\frac{1}{2} {\frac{1}{u} } \, du[/tex]
Step 3: Evaluate Further
- Integrate: [tex]-(ln|u|)|\limits^1_\frac{1}{2}[/tex]
- FTC Part 1: [tex]-(ln|1|-ln|\frac{1}{2} |)[/tex]
- Evaluate: [tex]-ln|2|[/tex]
- Evaluate: [tex]-0.693147[/tex]
