Calculus - Integration by Parts: trigonometric functions and substitutions, Exercises of Mathematics

Download Calculus - Integration by Parts: trigonometric functions and substitutions and more Exercises Mathematics in PDF only on Docsity! Page 1 Integration by parts; Integrating Trigonometric Functions; Trigonometric Substitutions 𝒅 𝒅𝒙 𝒔𝒊𝒏(𝒙) = 𝒄𝒐𝒔(𝒙) 𝒅 𝒅𝒙 𝒔𝒆𝒄(𝒙) = 𝒔𝒆𝒄(𝒙)𝒕𝒂𝒏(𝒙) 𝒅 𝒅𝒙 𝒄𝒐𝒔(𝒙) = −𝒔𝒊𝒏(𝒙) 𝒅 𝒅𝒙 𝒄𝒔𝒄(𝒙) = −𝒄𝒐𝒕(𝒙)𝒄𝒔𝒄(𝒙) 𝒅 𝒅𝒙 𝒕𝒂𝒏(𝒙) = 𝒔𝒆𝒄𝟐(𝒙) 𝒅 𝒅𝒙 𝒄𝒐𝒕(𝒙) = −𝒄𝒔𝒄𝟐(𝒙) Integrals: antiderivatives and area under a curve Page 2 Fundamental theorem of calculus: use to calculate value of definite integrals ∫ 𝒇′(𝒙)𝒅𝒙 = 𝒇(𝒃) − 𝒇(𝒂) 𝒃 𝒂 ∫ 𝒔𝒊𝒏(𝒙)𝒅𝒙 = − 𝒄𝒐𝒔(𝒙) + 𝑪 ∫ 𝒄𝒐𝒔(𝒙)𝒅𝒙 = 𝒔𝒊𝒏(𝒙) + 𝑪 ∫ 𝒕𝒂𝒏(𝒙)𝒅𝒙 = ∫ 𝒔𝒊𝒏(𝒙) 𝒄𝒐𝒔(𝒙) 𝒅𝒙 Use integration by substitution: Page 5 ∫ 𝒄𝒐𝒔(𝒙)(𝟐𝒙) 𝒅𝒙 = (𝟐𝒙)𝒔𝒊𝒏(𝒙) − ∫ 𝒔𝒊𝒏(𝒙) 𝟐𝒅𝒙 = (𝟐𝒙)𝒔𝒊𝒏(𝒙) + 𝟐𝒄𝒐𝒔(𝒙) = −(𝒙𝟐)(𝒄𝒐𝒔(𝒙)) + (𝟐𝒙)𝒔𝒊𝒏(𝒙) + 𝟐𝒄𝒐𝒔(𝒙) + 𝑪 = 𝒄𝒐𝒔(𝒙)(𝟐 − 𝒙𝟐) + 𝟐𝒙𝒔𝒊𝒏(𝒙) + 𝑪 Tabular integration Two functions multiplied together One of the function is a polynomial (since their derivatives go 0) Other function is easy to take antiderivative (often exponential or trig function) ∫ 𝒙𝟐𝒔𝒊𝒏(𝒙)𝒅𝒙 𝒙𝟐 + 𝒔𝒊𝒏(𝒙) 𝟐𝒙 - −𝒄𝒐𝒔(𝒙) 𝟐 + −𝒔𝒊𝒏(𝒙) 𝟎 𝒄𝒐𝒔(𝒙) −𝒙𝟐𝒄𝒐𝒔(𝒙) + 𝟐𝒙𝒔𝒊𝒏(𝒙) + 𝟐𝒄𝒐𝒔(𝒙) + 𝑪 Page 6 ∫(𝒙𝟐 + 𝟏)(𝒆−𝒙)𝒅𝒙 𝒙𝟐 + 𝟏 + 𝒆−𝒙 𝟐𝒙 - −𝒆−𝒙 𝟐 + 𝒆−𝒙 𝟎 −𝒆−𝒙 ∫(𝒙𝟐 + 𝟏)(𝒆−𝒙)𝒅𝒙 = (−𝒆−𝒙)(𝒙𝟐 + 𝟏) − (𝒆−𝒙)(𝟐𝒙) − 𝒆−𝒙(𝟐) + 𝑪 = (−𝒆−𝒙)(𝒙𝟐 + 𝟐𝒙 + 𝟑) + 𝑪 Trigonometry Integrals Reduction formulas: ∫ 𝒄𝒐𝒔𝒎(𝒙) 𝒔𝒊𝒏𝒏(𝒙)𝒅𝒙 = − 𝒄𝒐𝒔𝒎+𝟏(𝒙)𝒔𝒊𝒏𝒏−𝟏(𝒙) 𝒎 + 𝒏 + 𝒏 − 𝟏 𝒎 + 𝒏 ∫ 𝒄𝒐𝒔𝒎(𝒙) 𝒔𝒊𝒏𝒏−𝟐(𝒙)𝒅𝒙 ∫ 𝒕𝒂𝒏𝒎(𝒙)𝒔𝒆𝒄𝟐𝒌(𝒙)𝒅𝒙 = ∫ 𝒕𝒂𝒏𝒎(𝒙)𝒔𝒆𝒄𝟐(𝒙)𝒌−𝟏𝒔𝒆𝒄𝟐(𝒙)𝒅𝒙 = ∫ 𝒕𝒂𝒏𝒎(𝒙)(𝟏 + 𝒕𝒂𝒏𝟐𝒙)𝒌−𝟏𝒔𝒆𝒄𝟐(𝒙)𝒅𝒙 Then substitute u = tan(x) Page 7 (secant power is an even number) ∫ 𝒕𝒂𝒏𝟐𝒌+𝟏(𝒙)𝒔𝒆𝒄𝒏(𝒙)𝒅𝒙 = ∫ 𝒕𝒂𝒏𝟐(𝒙)𝒌𝒔𝒆𝒄𝒏−𝟏(𝒙)𝒔𝒆𝒄(𝒙)𝒕𝒂𝒏(𝒙)𝒅𝒙 = ∫(𝒔𝒆𝒄𝟐(𝒙) − 𝟏)𝒌𝒔𝒆𝒄𝒏−𝟏(𝒙)𝒔𝒆𝒄(𝒙)𝒕𝒂𝒏(𝒙)𝒅𝒙 Then substitute u = sec(x) (secant power is an odd number) Examples ∫ 𝒔𝒊𝒏𝟑(𝒙)𝒄𝒐𝒔𝟒(𝒙)𝒅𝒙 𝒎 = 𝟒 𝒏 = 𝟑 − 𝒄𝒐𝒔𝟓(𝒙)𝒔𝒊𝒏𝟐(𝒙) 𝟒 + 𝟑 + 𝟑 − 𝟏 𝟒 + 𝟑 ∫ 𝒄𝒐𝒔𝟒(𝒙) 𝒔𝒊𝒏(𝒙)𝒅𝒙 𝒖 = 𝒄𝒐𝒔(𝒙) 𝒅𝒖 − 𝒔𝒊𝒏(𝒙)𝒅𝒙 𝒅𝒙 = 𝒅𝒖 −𝒔𝒊𝒏(𝒙) ∫ 𝒄𝒐𝒔𝟒(𝒙) 𝒔𝒊𝒏(𝒙)𝒅𝒙 = ∫(𝒖)𝟒 𝒔𝒊𝒏(𝒙) 𝒅𝒖 −𝒔𝒊𝒏(𝒙) = − ∫ 𝒖𝟒𝒅𝒙 = − 𝒖𝟓 𝟓 = − 𝒄𝒐𝒔𝟓(𝒙) 𝟓 − 𝒄𝒐𝒔𝟓(𝒙)𝒔𝒊𝒏𝟐(𝒙) 𝟒 + 𝟑 + 𝟑 − 𝟏 𝟒 + 𝟑 (− 𝒄𝒐𝒔𝟓(𝒙) 𝟓 ) + 𝑪 = 𝒄𝒐𝒔𝟓(𝒙)𝒔𝒊𝒏𝟐(𝒙) 𝟕 − 𝟐 𝟑𝟓 𝒄𝒐𝒔𝟓(𝒙) + 𝑪