Calculus Formula Sheet: Trig, Derivatives, Integrals

Formula Sheet Trigonometric Identities/Definitions 𝒔𝒊𝒏𝟐 𝒙 + 𝒄𝒐𝒔𝟐 𝒙 = 𝟏 𝒔𝒊𝒏 𝟐𝒙 = 𝟐 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙 𝒄𝒐𝒔 𝟐𝒙 = 𝒄𝒐𝒔𝟐 𝒙 − 𝒔𝒊𝒏𝟐 𝒙 = 𝟐𝒄𝒐𝒔𝟐 𝒙 − 𝟏 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙 𝟏 𝒔𝒆𝒄 𝒙 = 𝒄𝒐𝒔 𝒙 𝟏 𝒄𝒐𝒔 𝒙 = 𝒕𝒂𝒏 𝒙 𝒔𝒊𝒏 𝒙 𝟏 𝒄𝒔𝒄 𝒙 = 𝒔𝒊𝒏 𝒙 𝒕𝒂𝒏 𝒙 = 𝒄𝒐𝒕 𝒙 = 𝟏 + 𝒕𝒂𝒏𝟐 𝒙 = 𝒔𝒆𝒄𝟐 𝒙 𝟏 + 𝒄𝒐𝒕𝟐 𝒙 = 𝒄𝒔𝒄𝟐 𝒙 Elementary Derivatives/Differentiation Rules 𝒅 [𝒂𝒏𝒚 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕] = 𝟎 𝒅𝒙 𝒅 𝒏 [𝒙 ] = 𝒏𝒙𝒏−𝟏 𝒅𝒙 𝒅 𝟏 [√𝒙] = 𝒅𝒙 𝟐√ 𝒙 𝒅 [𝒔𝒊𝒏 𝒙] = 𝒄𝒐𝒔 𝒙 𝒅𝒙 𝒅 [𝒄𝒐𝒔 𝒙] = −𝒔𝒊𝒏 𝒙 𝒅𝒙 𝒅 [𝒕𝒂𝒏 𝒙] = 𝒔𝒆𝒄𝟐 𝒙 𝒅𝒙 𝒅 [𝒄𝒐𝒕 𝒙] = −𝒄𝒔𝒄𝟐 𝒙 𝒅𝒙 𝒅 [𝒔𝒆𝒄 𝒙] = 𝒔𝒆𝒄 𝒙 𝒕𝒂𝒏 𝒙 𝒅𝒙 𝒅 [𝒄𝒔𝒄 𝒙] = −𝒄𝒔𝒄 𝒙 𝒄𝒐𝒕 𝒙 𝒅𝒙 𝒅 𝟏 [𝒍𝒏 𝒙] = 𝒅𝒙 𝒙 𝒅 [𝒙𝒍𝒏 𝒙] = 𝒍𝒏 𝒙 + 𝟏 𝒅𝒙 𝒅 𝒙 [𝒆 ] = 𝒆𝒙 𝒅𝒙 Product Rule:(𝒖𝒗)′ = 𝒖′ 𝒗 + 𝒗′𝒖 𝒖 ′ Quotient Rule:(𝒗) = 𝒖′ 𝒗−𝒗′ 𝒖 𝒗𝟐 Chain Rule:[𝒇(𝒖)]′ = 𝒇′ (𝒖)𝒖′ Inverse Rule: [𝒇−𝟏 (𝒙)]′ = 𝟏 𝒇′ [𝒇−𝟏 (𝒙)] For any real number𝒂, 𝒖(𝒙) [∫ ′ 𝒇(𝒕)𝒅𝒕] = 𝒇[𝒖(𝒙)]𝒖′ (𝒙) 𝒂 Elementary Integrals 𝒙𝒏+𝟏 ∫ 𝒙 𝒅𝒙 = + 𝑪 𝒊𝒇 𝒏 ≠ −𝟏 𝒏+𝟏 𝒏 ∫ 𝟏 𝒅𝒙 = 𝒍𝒏|𝒙| + 𝑪 𝒙 ∫ 𝒔𝒊𝒏 𝒂𝒙 𝒅𝒙 = −(𝒄𝒐𝒔 𝒂𝒙)/𝒂 + 𝑪 ∫ 𝒄𝒐𝒔 𝒂𝒙 𝒅𝒙 = (𝒔𝒊𝒏 𝒂𝒙)/𝒂 + 𝑪 ∫ 𝒕𝒂𝒏 𝒙 𝒅𝒙 = −𝒍𝒏|𝒄𝒐𝒔 𝒙| + 𝑪 = 𝒍𝒏|𝒔𝒆𝒄 𝒙| + 𝑪 ∫ 𝒆𝒙 𝒅𝒙 = 𝒆𝒙 + 𝑪 ∫ 𝒔𝒆𝒄 𝒙 𝒅𝒙 = 𝒍𝒏|𝒔𝒆𝒄 𝒙 + 𝒕𝒂𝒏 𝒙| + 𝑪 General Exponential and Logarithms of Base a 𝒂𝒙 ≔ 𝒆𝒙𝒍𝒏 𝒂 𝒍𝒐𝒈𝒂 𝒙 ≔ 𝒍𝒏 𝒙 𝒍𝒏 𝒂 𝒅 𝒙 [𝒂 ] = 𝒂𝒙 𝒍𝒏 𝒂 𝒅𝒙 𝒅 𝟏 [𝒍𝒐𝒈𝒂 𝒙] = 𝒅𝒙 𝒙 𝒍𝒏 𝒂 ∫ 𝒂𝒙 𝒅𝒙 = ∫ 𝒂𝒙 +𝑪 𝒍𝒏 𝒂 𝟏 𝒅𝒙 = 𝒍𝒐𝒈𝒂 |𝒙| + 𝑪 𝒙 𝒍𝒏 𝒂 Inverse Trigonometric Functions Function Domain 𝐬𝐢𝐧−𝟏 𝒙 [−𝟏, 𝟏] 𝐜𝐨𝐬 −𝟏 𝒙 [−𝟏, 𝟏] Range 𝝅 𝝅 [− , ] 𝟐 𝟐 [𝟎, 𝝅] 𝝅 𝝅 [−∞, ∞] (− , ) 𝒕𝒂𝒏−𝟏 𝒙 𝟐 𝟐 𝟏 𝐬𝐞𝐜 −𝟏 𝒙 = 𝐜𝐨𝐬 −𝟏 ( ) 𝒙 Definition 𝒚 = 𝐬𝐢𝐧−𝟏 𝒙 means 𝐬𝐢𝐧 𝒚 = 𝒙 𝒚 = 𝐜𝐨𝐬 −𝟏 𝒙 means 𝐜𝐨𝐬 𝒚 = 𝒙 𝒚 = 𝐭𝐚𝐧−𝟏 𝒙 means 𝐭𝐚𝐧 𝒚 = 𝒙 𝟏 𝐜𝐬𝐜 −𝟏 𝒙 = 𝐬𝐢𝐧−𝟏 ( ) 𝒙 𝟏 𝐜𝐨𝐭 −𝟏 𝒙 = 𝐭𝐚𝐧−𝟏 ( ) 𝒙 Derivatives of Inverse Trigonometric Functions Trig. Function --- negate ---> 𝒅 𝟏 [𝐬𝐢𝐧−𝟏 𝒙] = 𝒅𝒙 √𝟏 − 𝒙𝟐 𝒅 𝟏 [𝐭𝐚𝐧−𝟏 𝒙] = 𝒅𝒙 𝟏 + 𝒙𝟐 𝒅 𝟏 [𝐬𝐞𝐜 −𝟏 𝒙] = 𝒅𝒙 |𝒙|√𝒙𝟐 − 𝟏 Co-Trig Function 𝒅 𝟏 [𝐜𝐨𝐬 −𝟏 𝒙] = − 𝒅𝒙 √𝟏 − 𝒙𝟐 𝒅 𝟏 [𝐜𝐨𝐭 −𝟏 𝒙] = − 𝒅𝒙 𝟏 + 𝒙𝟐 𝒅 𝟏 [𝐜𝐬𝐜 −𝟏 𝒙] = − 𝒅𝒙 |𝒙|√𝒙𝟐 − 𝟏 Integrals Evaluated with Inverse Trigonometric Functions ∫ 𝟏 𝒖 𝒅𝒖 = 𝐬𝐢𝐧−𝟏 ( ) + 𝑪 𝒂 √𝒂𝟐 − 𝒖𝟐 𝟏 𝒖 𝐭𝐚𝐧−𝟏 ( ) + 𝑪 𝒂 𝒂 𝟏 𝟏 𝒖 ∫ 𝒅𝒖 = 𝐬𝐞𝐜 −𝟏 | | + 𝑪 𝒂 𝒂 𝒖√𝒖𝟐 − 𝒂𝟐 ∫ 𝟏 𝒂𝟐 + 𝒖𝟐 𝒅𝒖 = Hyperbolic Functions Name Hyperbolic Sine Hyperbolic Cosine Definition 𝒆𝒙 − 𝒆−𝒙 𝐬𝐢𝐧𝐡 𝒙 = 𝟐 𝒙 𝒆 + 𝒆−𝒙 𝐜𝐨𝐬𝐡 𝒙 = 𝟐 Derivative Integral 𝒅 [𝒔𝒊𝒏𝒉 𝒙] = 𝐜𝐨𝐬𝐡 𝒙 𝒅𝒙 𝒅 [𝒄𝒐𝒔𝒉 𝒙] = 𝐬𝐢𝐧𝐡 𝒙 𝒅𝒙 ∫ 𝐬𝐢𝐧𝐡 𝒙 𝒅𝒙 = 𝐜𝐨𝐬𝐡 𝒙 + 𝑪 ∫ 𝐜𝐨𝐬𝐡 𝒙 𝒅𝒙 = 𝐬𝐢𝐧𝐡 𝒙 + 𝑪 Integration by Parts 𝒃 ∫ 𝒖 𝒅𝒗 = 𝒖𝒗 − ∫ 𝒗 𝒅𝒖 𝒃 ∫ 𝒖 𝒅𝒗 = [𝒖𝒗]𝒃𝒂 − ∫ 𝒗 𝒅𝒖 𝒂 𝒂 Priority for choosing 𝒖 is: Logarithmic, Inverse Trigonometric, Algebraic/Polynomial, Trigonometric, Exponential Integration of Trigonometric Functions ∫ 𝐬𝐢𝐧𝒎 𝒙 𝐜𝐨𝐬𝒏 𝒙 𝒅𝒙 ∫ 𝐬𝐢𝐧𝒎 𝒙 𝐜𝐨𝐬𝒏 𝒙 𝒅𝒙 If any of 𝒎 and 𝒏 is odd, then write it as (2 x something + 1) and replace 𝐬𝐢𝐧𝟐 𝒙 = 𝟏 − 𝐜𝐨𝐬𝟐 𝒙or 𝐜𝐨𝐬𝟐 𝒙 = 𝟏 − 𝒔𝒊𝒏𝟐 𝒙 and substitute 𝒖 = 𝐜𝐨𝐬 𝒙or𝒖 = 𝐬𝐢𝐧 𝒙, respectively. If both of 𝒎 and 𝒏 is even, then replace 𝟏+𝐜𝐨𝐬 𝟐𝒙 𝟏−𝐜𝐨𝐬 𝟐𝒙 𝐜𝐨𝐬𝟐 𝒙 = and/or 𝐬𝐢𝐧𝟐 𝒙 = 𝟐 𝟐 until an odd power appears, and so use the method above. 𝒌 Integrals involving square roots 𝒌 Replace𝟏 + 𝐜𝐨𝐬 𝒌𝒙 = 𝟐 𝐜𝐨𝐬𝟐 𝟐 𝒙or𝟏 − 𝐜𝐨𝐬 𝒌𝒙 = 𝟐 𝐬𝐢𝐧𝟐 𝟐 𝒙 Use 𝒔𝒊𝒏𝟐 𝒙 + 𝒄𝒐𝒔𝟐 𝒙 = 𝟏 Rationalize: Multiply with the conjugate 𝟏 ∫[𝒄𝒐𝒔(𝒂 − 𝒃)𝒙 − 𝒄𝒐𝒔(𝒂 + 𝒃)𝒙]𝒅𝒙 𝟐 𝟏 ∫ 𝒔𝒊𝒏 𝒂𝒙 𝒄𝒐𝒔 𝒃𝒙 𝒅𝒙 = ∫[𝒔𝒊𝒏(𝒂 − 𝒃)𝒙 + 𝒔𝒊𝒏(𝒂 + 𝒃)𝒙] 𝒅𝒙 𝟐 𝟏 ∫ 𝒄𝒐𝒔 𝒂𝒙 𝒄𝒐𝒔 𝒃𝒙 𝒅𝒙 = ∫[𝒄𝒐𝒔(𝒂 − 𝒃)𝒙 + 𝒄𝒐𝒔(𝒂 + 𝒃)𝒙] 𝒅𝒙 𝟐 Write the expression inside the integral as a product of 𝒕𝒂𝒏𝟐 𝒙 and 𝒔𝒆𝒄𝟐 𝒙 𝟐 Then replace 𝒕𝒂𝒏 𝒙 = 𝒔𝒆𝒄𝟐 𝒙 − 𝟏 or 𝒔𝒆𝒄𝟐 𝒙 = 𝒕𝒂𝒏𝟐 𝒙 + 𝟏 Finally, integrate by parts: Let 𝒅𝒗 = 𝒔𝒆𝒄𝟐 𝒙 𝒅𝒙 or 𝒔𝒆𝒄 𝒙 𝒕𝒂𝒏 𝒙 𝒅𝒙, or integrate by substitution: Let 𝒅𝒖 = 𝒔𝒆𝒄𝟐 𝒙 𝒅𝒙 or𝒔 𝒆𝒄 𝒙 𝒕𝒂𝒏 𝒙 𝒅𝒙 𝟏 𝟏 ∫ 𝒔𝒆𝒄𝟑 𝒙 𝒅𝒙 = 𝒔𝒆𝒄 𝒙 𝒕𝒂𝒏𝒙 + 𝒍𝒏|𝒔𝒆𝒄 𝒙 + 𝒕𝒂𝒏𝒙| + 𝑪 𝟐 𝟐 𝟏 𝒙 𝟏 ∫ 𝒄𝒐𝒔𝟐 𝒙 𝒅𝒙 = (𝒙 + 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 )𝒙 + 𝑪 = + 𝒔𝒊𝒏 𝟐𝒙 + 𝑪 𝟐 𝟐 𝟒 𝟏 𝒙 𝟏 ∫ 𝒔𝒊𝒏𝟐 𝒙 𝒅𝒙 = (𝒙 − 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 )𝒙 + 𝑪 = − 𝒔𝒊𝒏 𝟐𝒙 + 𝑪 𝟐 𝟐 𝟒 ∫ 𝒔𝒊𝒏 𝒂𝒙 𝒔𝒊𝒏 𝒃𝒙 𝒅𝒙 = Integrals involving products of sines and cosines Integrals involving products and powers of secants and tangents Very useful integrals Integration with Trigonometric Substitution If the integral contains √𝒂𝟐 − 𝒖𝟐 Substitute 𝒖 = 𝒂 𝒔𝒊𝒏 𝜽 √𝒂𝟐 + 𝒖𝟐 √𝒖𝟐 − 𝒂𝟐 𝒖 = 𝒂 𝒕𝒂𝒏 𝜽 𝒖 = 𝒂 𝒔𝒆𝒄 𝜽 Integration of Rational Functions with Partial Fractions 𝑷(𝒙) Consider ∫ 𝑸(𝒙) 𝒅𝒙 where P(x) and Q(x) are polynomials. 1) If deg P(x) < deg Q(x) and if then use the decomposition 𝑷(𝒙) 𝑨𝟏 𝑨𝟐 𝑨𝒏 = + + ⋯+ 𝑸(𝒙) 𝒙 − 𝒙𝟏 𝒙 − 𝒙𝟐 𝒙 − 𝒙𝒏 and the Heaviside Cover-up Method to find 𝑨𝟏 , 𝑨𝟐 , … , 𝑨𝒏 𝑷(𝒙) 𝑨𝟏 𝑨𝟐 𝑨𝒏 = + + ⋯+ 𝟐 (𝒙 − 𝒓)𝒏 𝑸(𝒙) 𝒙 − 𝒓 (𝒙 − 𝒓) 𝑸(𝒙) = (𝒙 − 𝒙𝟏 )(𝒙 − 𝒙𝟐 ) … (𝒙 − 𝒙𝒏 ) are distinct linear factors 𝑸(𝒙) = (𝒙 − 𝒓)𝒏 are repeated linear factors 𝑸(𝒙) = (𝒑𝒙𝟐 + 𝒒𝒙 + 𝒓)𝒏 are repeated irreducible quadratic factors 𝑸(𝒙) = product of linear and irreducible quadratic factors 𝑷(𝒙) 𝑨𝟏 𝒙 + 𝑩𝟏 𝑨𝟐 𝒙 + 𝑩𝟐 𝑨𝒏 𝒙 + 𝑩𝒏 = 𝟐 + + ⋯+ 𝟐 𝟐 (𝒑𝒙𝟐 + 𝒒𝒙 + 𝒓)𝒏 𝑸(𝒙) 𝒑𝒙 + 𝒒𝒙 + 𝒓 (𝒑𝒙 + 𝒒𝒙 + 𝒓) 𝑷(𝒙) 𝑸(𝒙) = sum of decompositions from corresponding cases 2) If deg P(x) ≥ deg Q(x), then use long division to write 𝑷(𝒙) 𝑹(𝒙) = 𝑺(𝒙) + 𝑸(𝒙) 𝑸(𝒙) where deg R(x) < deg Q(x). Integrate S(x) normally and use part one above for R(x)/Q(. The Heaviside Cover-up Method: A quick method for finding the coefficients 𝑨𝟏 , 𝑨𝟐 , … , 𝑨𝒏 in 𝑷(𝒙) 𝑨𝟏 𝑨𝟐 𝑨𝒏 = + +⋯+ (𝒙 − 𝒙𝟏 )(𝒙 − 𝒙𝟐 ) … (𝒙 − 𝒙𝒏 ) 𝒙 − 𝒙𝟏 𝒙 − 𝒙𝟐 𝒙 − 𝒙𝒏 To find 𝑨𝒌 : 1) Cover the factor 𝒙 − 𝒙𝒌 on the left-hand-side of the equation above 2) Evaluate the left-hand-side for 𝒙 = 𝒙𝒌 That is, 𝑨𝒌 = 𝑷(𝒙𝒌 ) (𝒙 − 𝒙𝟏 ) … (𝒙 − 𝒙𝒌−𝟏 )(𝒙 − 𝒙𝒌+𝟏 ) … (𝒙 − 𝒙𝒏 ) Improper Integrals ∞ If 𝒇(𝒙) is continuous on [𝒂, ∞), then 𝒃 ∫ 𝒇(𝒙)𝒅𝒙 ∶= 𝐥𝐢𝐦 ∫ 𝒇(𝒙)𝒅𝒙 𝒃→∞ 𝒂 𝒂 𝒃 Type I If 𝒇(𝒙) is continuous on(−∞, 𝒃], then 𝒃 ∫ 𝒇(𝒙)𝒅𝒙 ∶= 𝐥𝐢𝐦 ∫ 𝒇(𝒙)𝒅𝒙 𝒂→−∞ 𝒂 −∞ ∞ If 𝒇(𝒙) is continuous on (−∞, ∞), then 𝒂 ∞ ∫ 𝒇(𝒙)𝒅𝒙 ∶= ∫ 𝒇(𝒙)𝒅𝒙 + ∫ 𝒇(𝒙)𝒅𝒙 −∞ −∞ 𝒂 𝒃 If 𝒇(𝒙) is continuous on [𝒂, 𝒃), then ∫ 𝒇(𝒙)𝒅𝒙 ∶= 𝐥𝐢𝐦− ∫ 𝒇(𝒙)𝒅𝒙 𝒂 𝒃 Type II If 𝒇(𝒙) is continuous on (𝒂, 𝒃], then 𝒄 𝒄→𝒃 𝒂 𝒃 ∫ 𝒇(𝒙)𝒅𝒙 ∶= 𝐥𝐢𝐦+ ∫ 𝒇(𝒙)𝒅𝒙 𝒂 𝒄→𝒂 𝒄 𝒄 If 𝒇(𝒙) is continuous on [𝒂, 𝒃) ∪ (𝒃, 𝒄], then 𝒃 𝒄 ∫ 𝒇(𝒙)𝒅𝒙 ∶= ∫ 𝒇(𝒙)𝒅𝒙 + ∫ 𝒇(𝒙)𝒅𝒙 𝒂 𝒂 𝒃 Integral Convergence Tests The p-Integral Test on ∞ 𝟏 ∫ 𝒑 𝒅𝒙 𝒂 𝒙 where 𝒂 is any positive real number The Direct Comparison Test (DCT): If 𝒇(𝒙) and 𝒈(𝒙) are continuous on [𝒂, ∞) such that 𝟎 ≤ 𝒇(𝒙) ≤ 𝒈(𝒙) then ∞ Converges if 𝒑 > 1, and Diverges if𝒑 ≤ 𝟏 ∞ ∞ ∞ If ∫𝒂 𝒇(𝒙)𝒅𝒙 diverges, then ∫𝒂 𝒈(𝒙)𝒅𝒙 diverges (Smaller one diverges => larger one diverges) ∞ 𝟎 ≤ ∫ 𝒇(𝒙)𝒅𝒙 ≤ ∫ 𝒈(𝒙)𝒅𝒙 𝒂 ∞ If ∫𝒂 𝒈(𝒙)𝒅𝒙 converges, then ∫𝒂 𝒇(𝒙)𝒅𝒙 converges (Larger one converges => smaller one converges) 𝒂 𝟏 →DCT Tricks Compare with𝒙𝒑 by removing constants from the numerator or denominator, or using the following inequalities: 𝒔𝒊𝒏(𝒙) ≤ 𝟏, 𝒄𝒐𝒔(𝒙) ≤ 𝟏, 𝒍𝒏(𝒙) ≤ 𝒙, 𝒍𝒏(𝒙) ≤ √𝒙 The Limit Comparison Test (LCT): If 𝒇(𝒙) and 𝒈(𝒙) are continuous on [𝒂, ∞) such that 𝒇(𝒙) 𝐥𝐢𝐦 𝒙→∞ 𝒈(𝒙) is a positive real number, then →LCT Tricks ∞ ∞ ∫𝒂 𝒇(𝒙)𝒅𝒙 and ∫𝒂 𝒈(𝒙)𝒅𝒙 both converge or both diverge (both integrals behave likewise) 𝒏(𝒙) If 𝒇(𝒙) = 𝒅(𝒙), then compare with 𝑻𝒆𝒓𝒎 𝒘𝒊𝒕𝒉 𝒉𝒊𝒈𝒉𝒆𝒔𝒕 𝒈𝒓𝒐𝒘𝒕𝒉 𝒊𝒏 𝒏(𝒙) 𝒈(𝒙) = 𝑻𝒆𝒓𝒎 𝒘𝒊𝒕𝒉 𝒉𝒊𝒈𝒉𝒆𝒔𝒕 𝒈𝒓𝒐𝒘𝒕𝒉 𝒊𝒏 𝒅(𝒙) 𝟏 Order of growths:𝒆−𝒙^𝟐 , 𝒆−𝟐𝒙 , 𝒆−𝒙 , 𝒙−𝟑 , 𝒙−𝟐 , 𝒙−𝟏 , 𝒙−𝟏/𝟐 , 𝒍𝒏(𝒙) , 𝟏, 𝒍𝒏(𝒙), √𝒙, 𝒙, 𝒙𝟐 , 𝒙𝟑 , 𝒆𝒙 , 𝒆𝟐𝒙 , 𝒆𝒙 𝟐 slower <------------------------------------grows----------------------------------------->faster Sequences 𝒍𝒏 𝒏 =𝟎 𝒏→∞ 𝒏 𝐥𝐢𝐦 𝒏 𝐥𝐢𝐦 √𝒏 = 𝟏 𝒏→∞ 𝒙𝒏 = 𝟎 (𝒂𝒏𝒚 𝒙) 𝒏→∞ 𝒏! 𝒏 𝐥𝐢𝐦 √𝒙 = 𝟏 (𝒙 > 0) 𝒏→∞ 𝒙 𝒏 𝐥𝐢𝐦 (𝟏 + ) = 𝒆𝒙 𝒏→∞ 𝒏 𝒏! =𝟎 𝒏→∞ 𝒏𝒏 𝐥𝐢𝐦 𝐥𝐢𝐦 𝐥𝐢𝐦 𝒙𝒏 = 𝟎 (|𝒙| < 1) 𝐥𝐢𝐦 𝒙𝒏 = ∞ (𝒙 > 1) 𝒏→∞ 𝒏→∞ Infinite Series 𝒂𝟏 , 𝒂𝟐 , 𝒂𝟑 , 𝒂𝟒 , … ∞ ∞ ∑ 𝒂𝒏 = 𝐥𝐢𝐦 ∑ 𝒂𝒏 = 𝐥𝐢𝐦 𝑺𝑵 ∑ 𝒂𝒏 = 𝒂𝟏 + 𝒂𝟐 + 𝒂𝟑 + ⋯ 𝒏=𝟏 Sequence Infinite series 𝒏=𝟏 𝑵 𝑵→∞ 𝒏=𝟏 𝑵→∞ (used to evaluate a Telescoping series) 𝑵 Partial Sums 𝑺𝑵 = 𝒂𝟏 + 𝒂𝟐 + 𝒂𝟑 + ⋯ + 𝒂𝑵 = ∑ 𝒂𝒏 𝒏=𝟏 𝒂 ∞ ∑ 𝒂𝒓 Geometric Series 𝒏−𝟏 𝟐 Converges to 𝟏−𝒓 if |𝒓| < 1 𝟑 = 𝒂 + 𝒂𝒓 + 𝒂𝒓 + 𝒂𝒓 + ⋯ Diverges if |𝒓| ≥ 𝟏 𝒏=𝟏 Types of Series Convergence ∞ ∑∞ 𝑛=𝑘 𝑎𝑛 Converges Absolutely means that ∑𝑛=𝑘|𝑎𝑛 | converges. ∞ ∞ ∑∞ 𝑛=𝑘 𝑎𝑛 Converges Conditionally means that ∑𝑛=𝑘|𝑎𝑛 | diverges but ∑𝑛=𝑘 𝑎𝑛 converges. Convergence/Divergence Tests on ∑∞ 𝒏=𝒌 𝒂𝒏 Test Description Usage Nth-term If 𝐥𝐢𝐦 𝒂𝒏 ≠ 𝟎, then the series diverges Integral ∑∞ 𝒏=𝒌 𝒂𝒏 converges ↔ ∫𝒌 𝒂𝒙 𝒅𝒙 converges provided that 𝒂𝒙 is continuous, positive and decreasing on [𝒌, ∞) p-Series ∑∞ 𝒏=𝒌 𝒑 converges if 𝒑 > 1, and diverges if 𝒑 ≤ 𝟏, provided 𝒌 ≥ 𝟏. 𝒏→∞ ∞ Direct Comparison (DCT) Limit Comparison (LCT) 𝟏 Only to prove divergence It can evaluate the improper integral With DCT and LCT 𝒏 If you can bound it ∞ ∞ Given ∑∞ 𝒏=𝒌 𝒅𝒏 ≤ ∑𝒏=𝒌 𝒂𝒏 ≤ ∑𝒏=𝒌 𝒄𝒏 such that 𝒂𝒏 , 𝒄𝒏 , 𝒅𝒏 ≥ 𝟎 after some above/below with index 𝒌, then: as series that ∞ (i) If ∑∞ 𝒏=𝒌 𝒄𝒏 converges, then ∑𝒏=𝒌 𝒂𝒏 converges converges/diverge ∞ (ii) If ∑∞ 𝒏=𝒌 𝒅𝒏 diverges, then ∑𝒏=𝒌 𝒂𝒏 diverges s, respectively Suppose that 𝒂𝒏 , 𝒃𝒏 > 0 after some index 𝒌, then: If 𝒂𝒏 is a division of 𝒂𝒏 two functions of n (i) If 𝐥𝐢𝐦 𝒃 ∈ (0, ∞), 𝒏→∞ 𝒏 ∞ then ∑∞ 𝒏=𝒌 𝒂𝒏 and ∑𝒏=𝒌 𝒃𝒏 both converge or both diverge 𝒂 ∞ (ii) If 𝐥𝐢𝐦 𝒃𝒏 = 𝟎 and ∑∞ 𝒏=𝒌 𝒃𝒏 converges, then ∑𝒏=𝒌 𝒂𝒏 converges 𝒏→∞ 𝒏 𝒂𝒏 ∞ (iii) If 𝐥𝐢𝐦 𝒃 = ∞ and ∑∞ 𝒏=𝒌 𝒃𝒏 diverges, then ∑𝒏=𝒌 𝒂𝒏 diverges 𝒏→∞ 𝒏 𝒂 Ratio 𝐂𝐨𝐧𝐯𝐞𝐫𝐠𝐞𝐬 𝐢𝐟 𝐥𝐢𝐦 𝒂𝒏+𝟏 < 1 𝒏→∞ 𝒏 ∞ The series ∑𝒏=𝒌 𝒂𝒏 , with 𝒂𝒏 > 0 { 𝒂𝒏+𝟏 𝐃𝐢𝐯𝐞𝐫𝐠𝐞𝐬 𝐢𝐟 𝐥𝐢𝐦 𝒂 > 1 𝒏→∞ Root The series ∑∞ 𝒏=𝒌 𝒂𝒏 , with 𝒂𝒏 ≥ 𝟎 { 𝒏 𝐂𝐨𝐧𝐯𝐞𝐫𝐠𝐞𝐬 𝐢𝐟 𝐥𝐢𝐦 𝒏√𝒂𝒏 < 1 𝒏→∞ 𝐃𝐢𝐯𝐞𝐫𝐠𝐞𝐬 𝐢𝐟 𝐥𝐢𝐦 𝒏√𝒂𝒏 > 1 𝒏→∞ Alternating Series (AST) 𝒏 The series ∑∞ 𝒏=𝒌(−𝟏) 𝒂𝒏 converges if all three conditions below are satisfied: (i) All 𝒂𝒏 ’s are positive (ii) The sequence {𝒂𝒏 } is non-increasing after some index (iii) 𝐥𝐢𝐦 𝒂𝒏 = 𝟎 𝒏→∞ If 𝒂𝒏 is a product of terms like factorials and powers If 𝒂𝒏 is an nth power of some terms, but not necessarily all Only to prove convergence when the sign changes from every term to the next If it is easier to show absolute convergence Absolute Absolute Convergence implies Convergence: Convergence ∞ If ∑∞ 𝒏=𝒌|𝒂𝒏 | Converges, then ∑𝒏=𝒌 𝒂𝒏 Converges (ACT) 𝒏 To find the Interval of Convergence (IOC) of a Power Series ∑∞ 𝒏=𝟎 𝒄𝒏 (𝒙 − 𝒂) : (i) (ii) Let𝒂𝒏 = 𝒄𝒏 (𝒙 − 𝒂)𝒏 , so that |𝒂𝒏 | = |𝒄𝒏 ||𝒙 − 𝒂|𝒏 Use the Ratio or Root test to find an open interval (𝒔, 𝒕) on which the power series converges |𝒂 | 𝒏 absolutely by forcing 𝐥𝐢𝐦 |𝒂𝒏+𝟏| < 1or 𝐥𝐢𝐦 √|𝒂𝒏 | < 1. 𝒏→∞ (iii) (iv) 𝒏 𝒏→∞ Now check if the power series converges (conditionally) or diverges at the endpoints 𝒙 = 𝒔, 𝒕. Finally, IOC= interval containing all x values on which the power series converges. Taylor & MacLaurin Series: The Taylor series of an infinitely differentiable function 𝒇(𝒙) centered at 𝒂 is given by ∞ 𝒇(𝒏) (𝒂) 𝒇′′ (𝒂) 𝒇′′′ (𝒂) 𝒏 ′ (𝒂)(𝒙 𝟐 (𝒙 (𝒙 (𝒙 − 𝒂)𝟑 + ⋯ ∑ − 𝒂) = 𝒇(𝒂) + 𝒇 − 𝒂) + − 𝒂) + 𝒏! 𝟐! 𝟑! 𝒏=𝟎 When 𝒂 = 𝟎, we obtain the MacLaurin series of 𝒇(𝒙) as ∞ ∑ 𝒏=𝟎 ∞ 𝒆𝒙 = ∑ 𝒏=𝟎 𝒇(𝒏) (𝟎) 𝒏 𝒇′′ (𝟎) 𝟐 𝒇′′′ (𝟎) 𝟑 𝒙 = 𝒇(𝟎) + 𝒇′ (𝟎)𝒙 + 𝒙 + 𝒙 +⋯ 𝒏! 𝟐! 𝟑! 𝒙𝒏 𝒙𝟐 𝒙𝟑 =𝟏+𝒙+ + +⋯ 𝒏! 𝟐! 𝟑! ∞ 𝒄𝒐𝒔 𝒙 = ∑ 𝒏=𝟎 (−𝟏)𝒏 𝒙𝟐𝒏 𝒙𝟐 𝒙𝟒 𝒙𝟔 =𝟏− + − +⋯ (𝟐𝒏)! 𝟐! 𝟒! 𝟔! ∞ 𝒔𝒊𝒏 𝒙 = ∑ 𝒏=𝟎 (−𝟏)𝒏 𝒙𝟐𝒏+𝟏 𝒙𝟑 𝒙𝟓 𝒙𝟕 =𝒙− + − +⋯ (𝟐𝒏 + 𝟏)! 𝟑! 𝟓! 𝟕! Polar  Rectangular 𝒙 = 𝒓 𝒄𝒐𝒔 𝜽 𝒚 = 𝒓 𝒔𝒊𝒏 𝜽 𝒓 𝟐 = 𝒙 𝟐 + 𝒚𝟐 𝒚 𝒚 𝒕𝒂𝒏 𝜽 = 𝐨𝐫 𝜽 = 𝒕𝒂𝒏−𝟏 𝒙 𝒙 These identities will be needed for symmetries: 𝒔𝒊𝒏(−𝜽) = −𝒔𝒊𝒏(𝜽) 𝒄𝒐𝒔(−𝜽) = 𝒄𝒐𝒔(𝜽) 𝒔𝒊𝒏(𝝅 − 𝜽) = 𝒔𝒊𝒏(𝜽) 𝒄𝒐𝒔(𝝅 − 𝜽) = −𝒄𝒐𝒔(𝜽) Symmetries of a Polar Curve 𝒓(𝜽) If the speed coefficient of  is 1,2,3,.. e.g. 𝒓(𝜽) = 𝒔𝒊𝒏(𝟐𝜽) If… Then symmetric w.r.t. x-axis 𝒓(−𝜽) = 𝒓(𝜽) y-axis 𝒓(−𝜽) = −𝒓(𝜽) y-axis 𝒓(𝝅 − 𝜽) = 𝒓(𝜽) x-axis 𝒓(𝝅 − 𝜽) = −𝒓(𝜽) origin 𝒓(𝝅 + 𝜽) = 𝒓(𝜽) If the speed coefficient of  is 1/k e.g. 𝒓(𝜽) = 𝒔𝒊𝒏(𝜽/𝟐); 𝒌 = 𝟐 If… Then symmetric w.r.t. x-axis 𝒓(−𝜽 + 𝟐𝒌𝝅) = 𝒓(𝜽) y-axis 𝒓(−𝜽 + 𝟐𝒌𝝅) = −𝒓(𝜽) y-axis 𝒓(𝝅 − 𝜽 + 𝟐𝒌𝝅) = 𝒓(𝜽) x-axis 𝒓(𝝅 − 𝜽 + 𝟐𝒌𝝅) = −𝒓(𝜽) origin 𝒓(𝝅 + 𝜽 + 𝟐𝒌𝝅) = 𝒓(𝜽) Areas and LengthsRelated to Polar Curves 𝟏 𝜷 The area inside a polar curve𝒓(𝜽) between 𝜶 ≤ 𝜽 ≤ 𝜷 is given by𝑨 = 𝟐 ∫𝜶 𝒓𝟐 (𝜽) 𝒅𝜽 The area between two polar curves 𝒓𝟏 (𝜽)(inner) and 𝒓𝟐 (𝜽) (outer) between 𝜶 ≤ 𝜽 ≤ 𝜷 is given by 𝑨= 𝟏 𝜷 𝟐 ∫ [𝒓 (𝜽) − 𝒓𝟐𝟏 (𝜽)] 𝒅𝜽 𝟐 𝜶 𝟐 𝜷 The length of a polar curve𝒓(𝜽)where 𝜽 runs from 𝜶 to 𝜷 is given by𝑳 = ∫𝜶 √[𝒓(𝜽)]𝟐 + [𝒓′ (𝜽)]𝟐 𝒅𝜽

Calculus Formula Sheet: Trig, Derivatives, Integrals

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