Evaluate the definite integral. 1/4 csc(2πt) cot(2πt) dt 1/12

Accepted Solution

Answer:[tex]\displaystyle \int\limits^{\frac{1}{4}}_{\frac{1}{12}} {\csc (2\pi t) \cot (2\pi t)} \, dt = \frac{1}{2\pi}[/tex]General Formulas and Concepts:CalculusDifferentiationDerivativesDerivative NotationDerivative Property [Multiplied Constant]:                                                          [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]Basic Power Rule:f(x) = cxⁿf’(x) = c·nxⁿ⁻¹IntegrationIntegralsIntegration 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]U-SubstitutionStep-by-step explanation:Step 1: DefineIdentify[tex]\displaystyle \int\limits^{\frac{1}{4}}_{\frac{1}{12}} {\csc (2\pi t) \cot (2\pi t)} \, dt[/tex]Step 2: Integrate Pt. 1Identify variables for u-substitution.Set u:                                                                                                             [tex]\displaystyle u = 2\pi t[/tex][u] Differentiate [Basic Power Rule, Derivative Properties]:                       [tex]\displaystyle du = 2\pi \ dt[/tex][Bounds] Switch:                                                                                           [tex]\displaystyle \left \{ {{x = \frac{1}{4} ,\ u = 2\pi (\frac{1}{4}) = \frac{\pi}{2}} \atop {x = \frac{1}{12}} ,\ u = 2\pi (\frac{1}{12}) = \frac{\pi}{6}} \right.[/tex]Step 3: integrate Pt. 2[Integral] Rewrite [Integration Property - Multiplied Constant]:                 [tex]\displaystyle \int\limits^{\frac{1}{4}}_{\frac{1}{12}} {\csc (2\pi t) \cot (2\pi t)} \, dt = \frac{1}{2\pi}\int\limits^{\frac{1}{4}}_{\frac{1}{12}} {2\pi \csc (2\pi t) \cot (2\pi t)} \, dt[/tex][Integral] U-Substitution:                                                                               [tex]\displaystyle \int\limits^{\frac{1}{4}}_{\frac{1}{12}} {\csc (2\pi t) \cot (2\pi t)} \, dt = \frac{1}{2\pi}\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{6}} {\csc (u) \cot (u)} \, du[/tex][Integral] Trigonometric Integration:                                                             [tex]\displaystyle \int\limits^{\frac{1}{4}}_{\frac{1}{12}} {\csc (2\pi t) \cot (2\pi t)} \, dt = \frac{1}{2\pi}[-\csc (u)] \bigg| \limits^{\frac{\pi}{2}}_{\frac{\pi}{6}}[/tex]Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:           [tex]\displaystyle \int\limits^{\frac{1}{4}}_{\frac{1}{12}} {\csc (2\pi t) \cot (2\pi t)} \, dt = \frac{1}{2\pi}(1)[/tex]Simplify:                                                                                                         [tex]\displaystyle \int\limits^{\frac{1}{4}}_{\frac{1}{12}} {\csc (2\pi t) \cot (2\pi t)} \, dt = \frac{1}{2\pi}[/tex]Topic: AP Calculus AB/BC (Calculus I/I + II)Unit: Integration