5.5 Sustitucion

# 5.5 Sustitucion - Section 5.5 Indefinite Integrals and the...

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Section 5.5 Indefinite Integrals and the Substitution Rule 323 value( q2 ); 71-80. Example CAS commands: : (assigned function and values for a, and b may vary) Mathematica For transcendental functions the FindRoot is needed instead of the Solve command. The Map command executes FindRoot over a set of initial guesses Initial guesses will vary as the functions vary. Clear[x, f, F] {a, b}= {0, 2 }; f[x_] = Sin[2x] Cos[x/3] 1 F[x_] = Integrate[f[t], {t, a, x}] Plot[{f[x], F[x]},{x, a, b}] x/.Map[FindRoot[F'[x]==0, {x, #}] &,{2, 3, 5, 6}] x/.Map[FindRoot[f'[x]==0, {x, #}] &,{1, 2, 4, 5, 6}] Slightly alter above commands for 75 - 80. Clear[x, f, F, u] a=0; f[x_] = x 2x 3 2 u[x_] = 1 x 2 F[x_] = Integrate[f[t], {t, a, u(x)}] x/.Map[FindRoot[F'[x]==0,{x, #}] &,{1, 2, 3, 4}] x/.Map[FindRoot[F''[x]==0,{x,#}] &,{1, 2, 3, 4}] After determining an appropriate value for b, the following can be entered b = 4; Plot[{F[x], {x, a, b}] 5.5 INDEFINTE INTEGRALS AND THE SUBSTITUTION RULE 1. Let u 3x du 3 dx du dx oe Ê oe Ê oe " 3 sin 3x dx sin u du cos u C cos 3x C ' ' oe oe oe " " " 3 3 3 2. Let u 2x du 4x dx du x dx oe Ê oe Ê oe # " 4 x sin 2x dx sin u du cos u C cos 2x C ' ' a b # # " " " oe oe oe 4 4 4 3. Let u 2t du 2 dt du dt oe Ê oe Ê oe " # sec 2t tan 2t dt sec u tan u du sec u C sec 2t C ' ' oe oe oe " " " # # # 4. Let u 1 cos du sin dt 2 du sin dt oe Ê oe Ê oe t t t 2 2 " # # 1 cos sin dt 2u du u C 1 cos C ' ' ˆ ‰ ˆ ˆ oe oe oe t t 2 2 t 3 3 # # # # \$ # \$ 5. Let u 7x 2 du 7 dx du dx oe Ê oe Ê oe " 7 28(7x 2) dx (28)u du 4u du u C (7x 2) C ' ' ' oe oe oe oe & & & % % " 7 6. Let u x du 4x dx du x dx oe " Ê oe Ê oe % \$ \$ " 4 x x 1 dx u du C x 1 C ' ' \$ % # % # \$ " " # # a b a b oe oe oe 4 1 1 u

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324 Chapter 5 Integration 7. Let u 1 r du 3r dr 3 du 9r dr oe Ê oe Ê oe \$ # # 3u du 3(2)u C 6 1 r C ' 9r dr 1 r È "Î# "Î# \$ "Î# oe oe oe ' a b 8. Let u y 4y 1 du 4y 8y dy 3 du 12 y 2y dy oe Ê oe Ê oe % # \$ \$ a b a b 12 y 4y 1 y 2y dy 3u du u C y 4y 1 C ' ' a b a b a b % # \$ # \$ % # # \$ oe oe oe 9. Let u x 1 du x dx du x dx oe Ê oe Ê oe \$Î# "Î# # 3 2 3 È x sin x 1 dx sin u du sin 2u C x 1 sin 2x 2 C ' ' È ˆ ˆ ˆ ˆ # \$Î# # \$Î# \$Î# # " " " oe oe oe 2 2 u 3 3 4 3 6 10. Let u du dx oe Ê oe " " x x cos dx cos u du cos u du sin 2u C sin C ' ' ' " " " " " # # # # x x 4 2x 4 x u 2 ˆ ‰ ˆ ˆ a b a b oe oe oe oe sin C oe " " 2x 4 x 2 ˆ ‰ 11. (a) Let u cot 2 du 2 csc 2 d du csc 2 d oe Ê oe Ê oe ) ) ) ) ) # # " # csc 2 cot 2 d u du C C cot 2 C ' ' # # " " " # # # ) ) ) ) oe oe oe oe Š u u 4 4 (b) Let u csc 2 du 2 csc 2 cot 2 d du csc 2 cot 2 d oe Ê oe Ê oe ) ) ) ) ) ) ) " # csc 2 cot 2 d u du C C csc 2 C ' ' # # " " " # # # ) ) ) ) oe oe oe oe Š u u 4 4 12. (a) Let u 5x 8 du 5 dx du dx oe Ê oe Ê oe " 5 du u du 2u C u C 5x 8 C ' ' ' dx 2 2 5x 8 5 5 5 5 5 u È È " " " " "Î# "Î# "Î# oe oe oe oe oe Š ˆ È (b) Let u 5x 8 du (5x 8) (5) dx du oe Ê oe Ê oe È " # "Î# 2 dx 5 5x 8 È du u C 5x 8 C ' ' dx 2 2 2 5x 8 5 5 5 È oe oe oe È 13. Let u 3 2s du 2 ds du ds oe Ê oe Ê oe " # 3 2s ds u du u du u C (3 2s) C ' ' ' È È ˆ ˆ ‰ ˆ oe oe oe oe " " " " # # # "Î# \$Î# \$Î# 2 3 3 14. Let u 2x 1 du 2 dx du dx oe Ê oe Ê oe " # (2x 1) dx u du u du C (2x 1) C ' ' ' oe oe oe oe \$ \$ \$ % " " " " # # # ˆ ˆ ‰ Š u 4 8 15. Let u 5s 4 du 5 ds du ds oe Ê oe Ê oe " 5 ds du u du 2u C 5s 4 C ' ' ' " " " " " "Î# "Î# È È 5s 4 u 5 5 5 5 2 oe oe oe oe ˆ ˆ ‰ ˆ È 16. Let u 2 x du dx du dx oe Ê oe Ê oe dx 3 u du 3 C C ' ' ' 3 u 3 (2 x) u 1 2 x 3( du) # oe oe oe oe Š 17. Let u 1 du 2 d du d oe Ê oe Ê oe ) ) ) ) ) # " # 1 d u du u du u C 1 C ' ' ' ) ) ) ) È
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