lec 1-2.pdf - CALCULUS 2 Lec 1-2 Indefinite Integral Antiderivatives Integration by substitution Introduction Derivatives of power logarithmic

lec 1-2.pdf - CALCULUS 2 Lec 1-2 Indefinite Integral...

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Unformatted text preview: CALCULUS 2 Lec. 1-2 Indefinite Integral Antiderivatives Integration by substitution Introduction Derivatives of power, logarithmic, exponential, trigonometric and hyperbolic functions () ′ () () ′ () −1 ln 1/ ln sin() cos sinh cosh cos − sin cosh sinh tan sec 2 tanh sech2 cot − csc 2 coth −csch2 sec sec tan − csc cot() sech csch − sech tanh csc − csch coth Derivatives of inverse trigonometric and inverse hyperbolic functions ′ () () sin−1 1 1− −1 cos −1 () 1 − 2 1 1 + 2 −1 1 + 2 1 tan−1 () cot −1 () sec −1 () csc −1 () −1 2 2 − 1 −1 2 − 1 ′ () () sinh 1 1 + 2 1 cosh−1 2 − 1 1 1 − 2 1 1 − 2 −1 tanh−1 coth−1 sech−1 csch−1 1 − 2 −1 1 + 2 Antiderivatives Definition A function is called an antiderivative of on an interval if ′ = () ∀ ∈ () (i) sin is an antiderivative of cos on ℜ (ii) cosh is an antiderivative of sinh on ℜ (iii) ln is an antiderivative of 1 on ℜ− 0 Notes: (i) Some functions have no antiderivatives. (ii) Some antiderivatives are not in the form elementary 2 functions such as , tan (iii) Finding some antiderivatives may be tedious such as 2 + 2 (iv) Antiderivative is not unique, for example − 3 , + 2 , are antiderivative of Theorem If and are two antiderivatives of on an interval , then = + where c is an arbitrary constant. Definition (Indefinite integral) The family of all antiderivatives of a function is called the indefinite integral of and is written in the form () = + the variable of integration the integrand the sign of integration () −1 −1 (i) 1 2 (ii) sech tanh = − sech + (iii) = − ln −2 = =− 1 = ln = 1 1 + = ln + Theorem (i) = () (ii) () = + () (i) (ii) + 2 = + 2 cos−1 − tan−1 = 2 cos −1 − tan−1 + 2 Property of definite integral If (i) and have antiderivatives on an interval , and (ii) 1 and 2 are constants, then 1 + 2 () = 1 + 2 () () − 2sin + 3 = − − cos 4 − 3/2 +3 3/2 3/2 = + cos + 2 − 4 ln + − 4 ln + Example (5) (i) 3 − (i) 1 = = = (ii) 2 = 1 2 2 9 3 9 3 3 3 2 6 − + − 6 ln − 6 ln 1 4 (ii) −3 + −3 1 − 3 3 + + tan sec −tan ∗ sec +tan sec −tan sec2 −sec tan sec2 −tan2 sec 2 − sec tan = = = tan − sec + tan sec +tan Integration by substitution 2 3 − 4 Example (6) 3 Let = − 4 → = 3 2 = 3 2 3 = = 2 9 = 3 − 4 1 3 3/2 1 3 = ∗ + 3/2 3/2 + sinh5 Example (7) = sinh4 cosh sinh Let = cosh = 2 2 −1 = = 2 cosh9/2 9 − 5/2 2 5/2 −1 2 sinh cosh = sinh = 9/2 9/2 cosh2 = cosh + − 7/2 − 2 3/2 + 1 1/2 1/2 + 5 4 cosh2 5 + 2 cosh + sin 2 1−cos Example (8) = 2 sin cos 1−cos Let = 1 − cos = =2 2(1−) 1/2 1/2 − = sin −1/2 − 1/2 =2 3/2 3/2 = 4 1 − cos − + 4 3 1 − cos 3 2 + 2 −1 Example (9) Let = = 1 2 −1 = =− 1 2 1− − tanh−1 + = − coth−1 + − tanh−1 ( ) + = − coth−1 ( ) + if if <1 >1 if if <0 >0 1 1/2 + 1/3 Example (10) Let = 1/6 = 6 5 3 2 + =6 2 =6 =6 6 = −+1 3 2 − +− 3 2 1/2 1/3 = 2 − 3 6 5 = 3 −1 =6 1 − +1 ( 3 +1)−1 +1 ln + 1 + + 1/6 − 6 ln 1/6 + 1 + If is an antiderivative of on an interval , then () ′ = () + Proof Let = () = ′ () ′ = = + = + () ′ () ′ () () () +1 , ≠ −1 +1 ln () , () ≠ 0 ′ () () ′ () () ln sin () ′ () − cos () cosh () ′ () sinh () ′ () 1 − 2 () sin−1 () Example (11) (i) tan (ii) sin cos sec − sin cos tan = =− = − ln cos + = ln sec + Similarly cot = ln sin + , tanh = ln cosh + , coth = ln sinh + (i) (ii) Similarly sec +tan sec +tan sec2 +sec tan sec +tan sec = sec ∗ = = ln sec + tan + csc = ln csc − cot + csch = ln csch − coth + Example (12) sech 1 cosh sech = = cosh cosh2 cosh −1 sinh = = tan 1+sinh2 1 Or sech = cosh 2 2 = − = − ∗ + + = 2 2 = 2 tan−1 + +1 Or + sech = −2 tan−1 − + (Try to prove) Example (13) (i) 5 4 +6 3 2 5 +3 4 (i) 5 4 +6 3 2 5 +3 4 (ii) = (+2) 3−ln +2 2(5 4 +6 3 ) 2 5 +3 4 5 4 1 2 = ln 2 + 3 (ii) (+2) 3−ln +2 = =− + 1/(+2) 3−ln +2 −1/(+2) 3−ln +2 = −2 3 − ln + 2 + Antiderivatives in the forms of inverse Trigonometric or inverse hyperbolic functions 1 2 − 2 1 sinh−1 / sinh−1 / 2 + 2 + cosh−1 / − cosh−1 −/ 1 2 − 2 1 2 + 2 1 2 − 2 1 tan−1 / 1 sec −1 1 tanh−1 / 1 coth−1 / 1 2 − 2 1 2 − 2 1 2 1 − sech−1 1 + 2 − csch−1 < < − <1 >1 ∈ − , − 0 ≠0 Example (14) (ii) (i) (ii) 1 4 2 +9 1 2 +4 1 4 2 +9 (i) 2 +4 4− 2 = −1 sinh 2 = 1 4 = ∗ 1 9 2 +4 1 (iii) 1 4 (iii) 1 − 5 2 −9 + 1 −1 tan 3/2 3/2 + 1 −1 2 = tan + 6 3 1 5 −1 − 2 = sin −9 2 4− 2 + 5 −1 tanh 3 3 + coth Example (15) coth 9+16 cosh2 9+16 cosh2 = = = sinh Let = 1 4 = ∗ 9+16 2 −1 3 4 = cosh sinh 9+16 (1+sinh2 cosh sinh 25+16 sinh2 ) = cosh 1 4 −1 csch 3 4 9 2 + 16 + = 1 4 sinh −1 − csch 3 3 + Completing squares + Remember that = Let = + 3 = 2 +4 = 2 + 2 + 2 2 +6+13 Example (16) +3 2 −9+13 2 +3 2 +4 = = 1 −1 tan 2 2 = + 1 +3 −1 = tan + 2 2 Example (17) sech2 (i) (i) tanh2 Let 1 = = (ii) 2 = = = −4 tanh +29 (ii) 7+− 2 = tanh = sech2 2 −4+29 −2 −1 sinh 5 7+− 2 − −3 −3 −1 sin 4 = + = = 2 −16 + −2 2 +25 tanh −2 −1 sinh 5 − 2 −6−7 = 16− −3 2 + +3 −1 Example (18) +3 −1 = = 1 2 = 1 2 = 1 2 ∗ +3 +3 2+6 2 +2−3 = 2+2 2 +2−3 2 + 1 2 +3 2 +2−3 2+2+4 2 +2−3 4 +1 2 −4 +1 −1 + 2 − 3 + 4 cosh 2 +1 −1 2 − 3 + 2 cosh + 2 2 = 2 + = + Finish See you next week ...
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