Question

If anyone could help that would be great! The topic is calculus, and substitution + integrals. I need help with the ones circled in red. I will award brainliest and +20 points!!

Answer 1

Answer:

$\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, dx = -ln(2)$

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]:                                     $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]:                                                         $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Trig Integration

Logarithmic Integration

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

1. Set u:                                                                                                             $\displaystyle u = 1 + cos(x)$
2. [u] Differentiate [Trig Derivative]:                                                                 $\displaystyle du = -sin(x) \ dx$
3. [Bounds of Integration] Change:                                                                 $\displaystyle [\frac{1}{2}, 1]$

Step 3: Integrate Pt. 2

1. [Integral] Rewrite [Integration Property - Multiplied Constant]:                 $\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, dx = -\int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{-sin \ x}{1 + cos \ x}} \, dx$
2. [Integral] U-Substitution:                                                                               $\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, dx = -\int\limits^{1}_{\frac{1}{2}} {\frac{1}{u}} \, du$
3. [Integral] Logarithmic Integration:                                                               $\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, dx = -(ln|u|) \bigg| \limits^{1}_{\frac{1}{2}}$
4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:           $\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, dx = -ln(2)$

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

Unit: Integration

Book: College Calculus 10e