Sec 17

eu du = eu + c. sen u du = − cos u + c. cos u du = sen u + c. tg u du = ln |sec u| + c. cotg u du = ln |sen u| + c. sec u du = ln |sec u + tg u| + c. cosec u du = ln |cosec u − cotg u| + c. sec u tg u du = sec u + c. cosec u cotg u du = −cosec u + c. sec2 u du = tg u + c. cosec2 u du = −cotg u + c. du 1 = a arc tg u + c. a u2 +a2 du u2 −a2 1 = 2a ln √ du u2 +a2 √ du u2 −a2 √ du = arc sen u + c, a a2 −u2 1 √ du = a arc sec u a u u2 −a2 Sejam u e v fun¸˜es deriv´veis de x e n conco a…

eu du = eu + c. sen u du = − cos u + c. cos u du = sen u + c. tg u du = ln |sec u| + c. cotg u du = ln |sen u| + c. sec u du = ln |sec u + tg u| + c. cosec u du = ln |cosec u − cotg u| + c. sec u tg u du = sec u + c. cosec u cotg u du = −cosec u + c. sec2 u du = tg u + c. cosec2 u du = −cotg u + c. du 1 = a arc tg u + c. a u2 +a2 du u2 −a2 1 = 2a ln √ du u2 +a2 √ du u2 −a2 √ du = arc sen u + c, a a2 −u2 1 √ du = a arc sec u a u u2 −a2 Sejam u e v fun¸˜es deriv´veis de x e n conco a…

eu du = eu + c. sen u du = − cos u + c. cos u du = sen u + c. tg u du = ln |sec u| + c. cotg u du = ln |sen u| + c. sec u du = ln |sec u + tg u| + c. cosec u du = ln |cosec u − cotg u| + c. sec u tg u du = sec u + c. cosec u cotg u du = −cosec u + c. sec2 u du = tg u + c. cosec2 u du = −cotg u + c. du 1 = a arc tg u + c. a u2 +a2 du u2 −a2 1 = 2a ln √ du u2 +a2 √ du u2 −a2 √ du = arc sen u + c, a a2 −u2 1 √ du = a arc sec u a u u2 −a2 Sejam u e v fun¸˜es deriv´veis de x e n conco a…

eu du = eu + c. sen u du = − cos u + c. cos u du = sen u + c. tg u du = ln |sec u| + c. cotg u du = ln |sen u| + c. sec u du = ln |sec u + tg u| + c. cosec u du = ln |cosec u − cotg u| + c. sec u tg u du = sec u + c. cosec u cotg u du = −cosec u + c. sec2 u du = tg u + c. cosec2 u du = −cotg u + c. du 1 = a arc tg u + c. a u2 +a2 du u2 −a2 1 = 2a ln √ du u2 +a2 √ du u2 −a2 √ du = arc sen u + c, a a2 −u2 1 √ du = a arc sec u a u u2 −a2 Sejam u e v fun¸˜es deriv´veis de x e n conco a…

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3 seno: ¨ ..... ¨¨ .... α β .. c b b a cos(α) = b a cos(β) = c a tangente: tg(α) = c b tg(β) = b c cotangente: ctg(α) = b c ctg(β) = c b secante: ¨ sen(β) = sec(α) = a b sec(β) = a c cossecante: a ¨ ¨¨ c a cosseno: ....¨ ¨¨.......... sen(α) = csc(α) = a c csc(β) = a b Figura 1.4: As razões trigonométricas. Razões trigonométricas de alguns ângulos…

9. y = sen u ⇒ y = u cos u. 10. y = cos u ⇒ y = −u sen u. 11. y = tg u ⇒ y = u sec2 u. 12. y = cotg u ⇒ y = −u cosec2 u. 13. y = sec u ⇒ y = u sec u tg u. 14. y = cosec u ⇒ y = −u cosec u cotg u. 15. y = arc sen u ⇒ y = √1u u2 . − 16. y = arc u cos u ⇒ y = √−−u2 . 1 u tg u ⇒ y = 1+u2 . cot g u ⇒ 1−u 2 . +u 17. y = arc 18. y = arc 19. y = arc sec u, |u| 1 u ⇒ y = |u|√u2 −1 , |u| > 1. 20. y = arc cosec u, |u| 1 −u ⇒ y = |u|√u2 −1 , |u| > 1. • 1. 2.…

eu du = eu + c. sen u du = − cos u + c. cos u du = sen u + c. tg u du = ln |sec u| + c. cotg u du = ln |sen u| + c. sec u du = ln |sec u + tg u| + c. cosec u du = ln |cosec u − cotg u| + c. sec u tg u du = sec u + c. cosec u cotg u du = −cosec u + c. sec2 u du = tg u + c. cosec2 u du = −cotg u + c. du 1 = a arc tg u + c. a u2 +a2 du u2 −a2 1 = 2a ln √ du u2 +a2 √ du u2 −a2 √ du = arc sen u + c, a a2 −u2 1 √ du = a arc sec u a u u2 −a2 Sejam u e v fun¸˜es deriv´veis de x e n conco a…

eu du = eu + c. sen u du = − cos u + c. cos u du = sen u + c. tg u du = ln |sec u| + c. cotg u du = ln |sen u| + c. sec u du = ln |sec u + tg u| + c. cosec u du = ln |cosec u − cotg u| + c. sec u tg u du = sec u + c. cosec u cotg u du = −cosec u + c. sec2 u du = tg u + c. cosec2 u du = −cotg u + c. du 1 = a arc tg u + c. a u2 +a2 du u2 −a2 1 = 2a ln √ du u2 +a2 √ du u2 −a2 √ du = arc sen u + c, a a2 −u2 1 √ du = a arc sec u a u u2 −a2 Sejam u e v fun¸˜es deriv´veis de x e n conco a…

eu du = eu + c. sen u du = − cos u + c. cos u du = sen u + c. tg u du = ln |sec u| + c. cotg u du = ln |sen u| + c. sec u du = ln |sec u + tg u| + c. cosec u du = ln |cosec u − cotg u| + c. sec u tg u du = sec u + c. cosec u cotg u du = −cosec u + c. sec2 u du = tg u + c. cosec2 u du = −cotg u + c. du 1 = a arc tg u + c. a u2 +a2 du u2 −a2 1 = 2a ln √ du u2 +a2 √ du u2 −a2 √ du = arc sen u + c, a a2 −u2 1 √ du = a arc sec u a u u2 −a2 Sejam u e v fun¸˜es deriv´veis de x e n conco a…