lec 3-pRO.pdf - CALCULUS 2 Lec 3 Indefinite Integral Integration by Parts Successive Reduction Formulae Integration by Parts If and are two

lec 3-pRO.pdf - CALCULUS 2 Lec 3 Indefinite Integral...

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Unformatted text preview: CALCULUS 2 Lec. 3 Indefinite Integral Integration by Parts Successive Reduction Formulae Integration by Parts If and are two differentiable functions on an interval , then () = ′ + ()′ () () = () = ′ + ()′ () ′ + ()′ () ()′ () = − ′ Using = () and = () , we get sec 2 Example(31): Let = = + - = sec 2 = tan = − = tan − tan = tan − ln sec + Try to evaluate: cosh , , 2 − 3 + 4 Example(32): = 2 − 3 + 4 = = 2 − 3 = = 2 − 3 + 4 − 2 − 3 Let = 2 − 3 = = 2 = = 2 − 3 + 4 − 2 − 3 − 2 = 2 − 3 + 4 − 2 − 3 + 2 + = 2 − 5 + 9 + Let Try to evaluate: , 2 cos Example(32b): 2 − 3 + 4 In this problem, a simple method (tabular form) could be used and derivatives and integrals 2 − 3 + 4 + 2 − 3 2 + 0 = 2 − 3 + 4 2 − 3 2 + = 2 − 5 + 9 + This method could be used to evaluate , if: () is a polynomial of any degree, and () is sin , cos , sinh( + ) , cosh( + ) , + + tan−1 Example(33): Let = tan−1 = 1 1+ 2 −1 = tan − = tan Try to evaluate: = −1 = 2 1+ 1 − ln 1 + 2 + c 2 sin−1 , sinh−1 , … , cos−1 , … , csch−1 , ln 2 + 2 Example(34): Let = 2 + 2 = 2 2 2 + 2 = 2 + 2 − = 2 + 2 − = 2 + 2 − 2 = 2 2 = = 2 2 + 2 2 +2 −2 2 + 2 2 + 2 − 2 −1 sinh 2 2 + 2 + + + 1 2 −1 2 2 = + + sinh + 2 cos Example(35): Let = = cos = − sin cos = + Let = sin = 1 sin = 1 = cos = cos sin = + − cos 2 cos sin 1+ 2 = + + 2 = 2 2 cos + sin + + Successive Reduction Example(36): Find a reduction formula for = sin , then use it to evaluate sin5 = sin = sin−1 sin = sin−1 = − 1 sin−2 cos = sin = − cos = −sin−1 cos + − 1 sin−2 cos 2 = −sin−1 cos + − 1 sin−2 1 − sin2 Example(36): continued = −sin−1 cos + − 1 sin−2 − sin = −sin−1 cos + − 1 −2 − = −sin−1 cos + − 1 −2 = sin5 = = = sin−1 cos − + −1 −2 sin4 cos 4 = 5 = − + 3 5 5 sin4 cos 4 sin2 cos 2 − + − + 1 5 5 3 3 sin4 cos 4 sin2 cos 8 − − + − cos 5 15 15 cos − 3 sin4 + 4 sin2 + 8 + 15 + Example(37): Find a reduction formula for = sec = sec −2 sec 2 = sec −2 = − 2 sec −3 sec tan = sec −2 tan − − 2 = sec 2 = tan sec −2 tan2 = sec −2 tan − − 2 sec −2 sec 2 − 1 = sec −2 tan − − 2 sec − sec −2 = sec −2 tan − − 2 − −2 − 1 = sec −2 tan + − 2 −2 = sec−2 tan −1 + −2 −1 −2 Example(38): Find a reduction formula for = 1 + 2 = −1 1 + 2 = −1 = 1 + 2 1 2 = − 1 −2 −1 1+ 2 2 3 1/2 = ∗ ∗ 1 + 2 3/2 3/2 −1 −2 1 + 2 3/2 3 2 3/2 3 −1 1+ −1 = − −2 1 + 2 1 + 2 3 2 3/2 3 −1 1+ −1 = − −2 + 1 + 2 3 2 3/2 3 −1 1+ −1 = − −2 + 3 3 3/2 −1 1+ 2 +2 −1 = − 3 3 3 −2 = = − −1 1+ 2 +2 3/2 − −1 +2 −2 Example(39): Find a reduction formula for , = ln = ln = ln , = = −1 1 +1 ln +1 , = = − +1 +1 ln +1 ln +1 +1 −1 − +1 ,−1 Notes: i) After n steps we get ,0 = ii) = ln = +1 +1 is a special case (m=0) + Example(40): Find a reduction formula for , = sin cos , = sin cos −1 cos = cos −1 = sin cos = − 1 cos −2 (− sin ) , = sin+1 cos−1 +1 , = sin+1 cos−1 +1 , = 1+ sin+1 cos−1 +1 −1 +1 , = , = = sin+1 +1 + −1 +1 sin+2 cos −2 + −1 +1 sin 1 − cos 2 cos −2 + −1 +1 ,−2 − , sin+1 cos−1 +1 + sin+1 cos−1 + −1 +1 ,−2 + −1 ,−2 + Example(41): Find a reduction formula for , = tan sec , = tan sec −2 sec 2 = sec −2 = − 2 sec −3 sec tan , = , = tan+1 sec−2 +1 tan+1 sec−2 − − −2 +1 −2 = tan sec 2 tan+1 = +1 tan+2 sec −2 tan sec 2 − 1 sec −2 +1 +1 tan+1 sec−2 −2 , = − − ,−2 +1 +1 , −2 tan+1 sec−2 −2 1+ = + +1 , +1 +1 ,−2 tan+1 sec−2 −2 , = +−1 + +−1 ,−2 Example(42): Try to find reduction formulae in the following cases: (i) = cosh (ii) = tan (iii) = (iv) = (v) = cos (vi) = cosh (vii) , = (viii) , = 2 − 2 coth csch cos cos Finish See you next week ...
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