Question

Use integration by parts to integrate sin2x between pi and 0

Answer 1

`\displaystyle \int\limits^0_(\pi) {\sin (2x)} \, dx = 0`

General Formulas and Concepts:

Calculus

Integration

• Integrals

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`

U-Substitution

Explanation:

Step 1: Define

Identify

`\displaystyle \int\limits^0_(\pi) {\sin (2x)} \, dx`

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

1. Set u:
   `\displaystyle u = 2x`
2. [u] Differentiate:
   `\displaystyle du = 2 \ dx`
3. [Bounds] Switch:
   `\displaystyle \left \{ {{x = 0 ,\ u = 2(0) = 0} \atop {x = \pi ,\ u = 2 \pi}} \right.`

Step 3: Integrate Pt. 2

1. [Integral] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle \int\limits^0_(\pi) {\sin (2x)} \, dx = (1)/(2) \int\limits^0_(\pi) {2 \sin (2x)} \, dx`
2. [Integral] U-Substitution:
   `\displaystyle \int\limits^0_(\pi) {\sin (2x)} \, dx = (1)/(2) \int\limits^0_(2 \pi) {\sin u} \, du`
3. Trigonometric Integration:
   `\displaystyle \int\limits^0_(\pi) {\sin (2x)} \, dx = (1)/(2)(-\cos u) \bigg| \limits^0_(2 \pi)`
4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
   `\displaystyle \int\limits^0_(\pi) {\sin (2x)} \, dx = (1)/(2)(0)`
5. Simplify:
   `\displaystyle \int\limits^0_(\pi) {\sin (2x)} \, dx = 0`

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

Unit: Integration