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# Integral Parsial.pdf

Integral parsial substitusi khusus dan trigonometri

Integral parsial substitusi khusus dan trigonometri

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### Integral Parsial.pdf

1. 1. TEKNIK PENGINTEGRALAN A. Substitusi B. “Khusus” C. Integral Parsial D. Substitusi Trigonometri Materi yang harus dikuasai sebelumnya: 1. Integral Tak Tentu 2. Sifat-Sifat Integral Tak Tentu 3. Derivatif Fungsi Trigonometri 4. Derivatif Fungsi Eksponen dan Fungsi Logaritma
2. 2. !"#\$%&'()*+"%,!)#&!%-"-.\$#&! Rumus-Rumus : !sin 𝑎𝑥 𝑑𝑥 = −1 𝑎 . cos 𝑎𝑥 + 𝑐 1) !cos 𝑎𝑥 𝑑𝑥 = 1 𝑎 . sin 𝑎𝑥 + 𝑐 2) !sec! 𝑎𝑥 𝑑𝑥 = 1 𝑎 . tan 𝑎𝑥 + 𝑐 3) !csc! 𝑎𝑥 𝑑𝑥 = −1 𝑎 . cot 𝑎𝑥 + 𝑐 4) !sec 𝑎𝑥 tan 𝑎 𝑥𝑑𝑥 = 1 𝑎 . sec 𝑎𝑥 + 𝑐 5) !csc 𝑎𝑥 cot 𝑎𝑥 𝑑𝑥 = −1 𝑎 . csc 𝑎𝑥 + 𝑐 6) 7a) !tan 𝑎𝑥 𝑑𝑥 = −1 𝑎 . ln(cos 𝑎𝑥) + 𝑐 7b) !tan 𝑎𝑥 𝑑𝑥 = 1 𝑎 . ln(sec 𝑎𝑥) + 𝑐 8) !cot 𝑎𝑥 𝑑𝑥 = 1 𝑎 . ln(sin 𝑎𝑥) + 𝑐 9) !sec 𝑎𝑥 𝑑𝑥 = 1 𝑎 . ln(sec 𝑎𝑥 + tan 𝑎𝑥) + 𝑐 10) !csc 𝑎𝑥 𝑑𝑥 = 1 𝑎 . ln(csc 𝑎𝑥 − cot 𝑎𝑥) + 𝑐
3. 3. Sebelum melanjutkan ke slide berikutnya, sebaiknya kamu berkonsentrasi penuh yaaa J J J GENTLE WARNING !!! Fasten your seatbelt … the road will be a little bit rough J J J
4. 4. C. Teknik Pengintegralan Parsial ∫ 𝑢 𝑑𝑣 = 𝑢. 𝑣 − ∫ 𝑣 𝑑𝑢 Contoh 1. ! 3𝑥 . sin 2𝑥 𝑑𝑥 = 𝑑𝑢 𝑑𝑥 = 3 𝑣 = ! sin 2𝑥 𝑑𝑥 3𝑥 . − 1 2 cos 2𝑥 𝑢 = 3𝑥 − ! − 1 2 cos 2𝑥 . 3𝑑𝑥 𝑑𝑢 = 3 𝑑𝑥 𝑣 = − 1 2 cos 2𝑥 = − 3 2 𝑥. cos 2𝑥 + ! 3 2 cos 2𝑥 𝑑𝑥 = − 3 2 𝑥. cos 2𝑥 + 3 4 sin 2𝑥 + 𝑐
5. 5. Contoh 2. !(2𝑥 − 3) . cos(4𝑥 + 1) 𝑑𝑥 = ∫ 𝑢 𝑑𝑣 = 𝑢. 𝑣 − ∫ 𝑣 𝑑𝑢 𝑑𝑢 𝑑𝑥 = 2 𝑣 = ! cos(4𝑥 + 1) 𝑑𝑥 (2𝑥 − 3) . 1 4 sin(4𝑥 + 1) 𝑢 = 2𝑥 − 3 − ! 1 4 sin(4𝑥 + 1) . 2𝑑𝑥 𝑑𝑢 = 2 𝑑𝑥 𝑣 = 1 4 sin(4𝑥 + 1) = 2𝑥 − 3 . 1 4 sin 4𝑥 + 1 − ! 1 2 sin(4𝑥 + 1) 𝑑𝑥 = 2𝑥 − 3 . ! " sin 4𝑥 + 1 + ! # cos(4𝑥 + 1) + 𝑐
6. 6. Contoh 3. ! 5𝑥\$ . cos 3𝑥 𝑑𝑥 = ∫ 𝑢 𝑑𝑣 = 𝑢. 𝑣 − ∫ 𝑣 𝑑𝑢 𝑑𝑢 𝑑𝑥 = 10𝑥 𝑣 = ! cos 3𝑥 𝑑𝑥 5𝑥\$. 1 3 sin 3𝑥 𝑢 = 5𝑥\$ − ! 1 3 sin 3𝑥 . 10𝑥 𝑑𝑥 𝑑𝑢 = 10𝑥 𝑑𝑥 𝑣 = 1 3 sin 3𝑥 = 5 3 𝑥\$. sin 3𝑥 − ! 10 3 𝑥 . sin 3𝑥 𝑑𝑥 Masih berbentuk integral parsial Sehingga harus dilakukan kembali cara integral parsial Kita lanjutkan ke slide berikutnya J J ( * )
7. 7. ∫ 𝑢 𝑑𝑣 = 𝑢. 𝑣 − ∫ 𝑣 𝑑𝑢 ! 10 3 𝑥 . sin 3𝑥 𝑑𝑥 = 𝑑𝑢 𝑑𝑥 = 10 3 𝑣 = ! sin 3𝑥 𝑑𝑥 10 3 𝑥 . − 1 3 cos 3𝑥 𝑢 = 10 3 𝑥 − ! − 1 3 cos 3𝑥 . 10 3 𝑑𝑥 𝑑𝑢 = 10 3 𝑑𝑥 𝑣 = − 1 3 cos 3𝑥 = − 10 9 𝑥. cos 3𝑥 + ! 10 9 cos 3𝑥 𝑑𝑥 = − 10 9 𝑥. cos 3𝑥 + 10 27 sin 3𝑥 ! 10 3 𝑥 . sin 3𝑥 𝑑𝑥 = − 10 9 𝑥. cos 3𝑥 + 10 27 sin 3𝑥 ( ** )
8. 8. ! 5𝑥\$ . cos 3𝑥 𝑑𝑥 = 5 4 𝑥\$. sin 3𝑥 − ! 10 3 𝑥 . sin 3𝑥 𝑑𝑥 ( * ) ! 10 3 𝑥 . sin 3𝑥 𝑑𝑥 = − 10 9 𝑥. cos 3𝑥 + 10 27 sin 3𝑥 ( ** ) Jika kita substitusikan persamaan ( ** ) ke persamaan ( * ), maka diperoleh ! 5𝑥\$ . cos 3𝑥 𝑑𝑥 = 5 4 𝑥\$. sin 3𝑥 − − 10 9 𝑥. cos 3𝑥 + 10 27 sin 3𝑥 = 5 4 𝑥\$. sin 3𝑥 + 10 9 𝑥. cos 3𝑥 − 10 27 sin 3𝑥
9. 9. Contoh 4. ! 15𝑥\$ . 2𝑥 + 3𝑑𝑥 = ∫ 𝑢 𝑑𝑣 = 𝑢. 𝑣 − ∫ 𝑣 𝑑𝑢 𝑑𝑢 𝑑𝑥 = 30𝑥 𝑣 = ! 2𝑥 + 3 𝑑𝑥 15𝑥\$. 1 3 (2𝑥 + 3) % \$ 𝑢 = 15𝑥\$ − ! 1 3 (2𝑥 + 3) % \$. 30𝑥 𝑑𝑥 𝑑𝑢 = 30𝑥 𝑑𝑥 𝑣 = ! % (2𝑥 + 3) ! " = 5𝑥\$. (2𝑥 + 3) % \$− ! 10𝑥 . (2𝑥 + 3) % \$𝑑𝑥 Masih berbentuk integral parsial Sehingga harus dilakukan kembali cara integral parsial Kita lanjutkan ke slide berikutnya J J ( * ) 𝑤 = 2𝑥 + 3 𝑑𝑤 𝑑𝑥 = 2 1 2 𝑑𝑤 = 𝑑𝑥 𝑣 = ! 𝑤 " ! . 1 2 𝑑𝑤
10. 10. ∫ 𝑢 𝑑𝑣 = 𝑢. 𝑣 − ∫ 𝑣 𝑑𝑢 ! 10𝑥 . (2𝑥 + 3) % \$𝑑𝑥 = 𝑑𝑢 𝑑𝑥 = 10 𝑣 = !(2𝑥 + 3) # ! 𝑑𝑥 10𝑥 . ! . (2𝑥 + 3) # " 𝑢 = 10𝑥 − ! 1 5 (2𝑥 + 3) . \$. 10𝑑𝑥 𝑑𝑢 = 10 𝑑𝑥 𝑣 = 1 5 (2𝑥 + 3) . \$ = 2𝑥. (2𝑥 + 3) . \$ − ! 2. (2𝑥 + 3) . \$ 𝑑𝑥 𝑤 = 2𝑥 + 3 𝑑𝑤 𝑑𝑥 = 2 1 2 𝑑𝑤 = 𝑑𝑥 𝑣 = ! 𝑤 # !. 1 2 𝑑𝑤 𝑤 = 2𝑥 + 3 𝑑𝑤 𝑑𝑥 = 2 1 2 𝑑𝑤 = 𝑑𝑥 = 2 7 (2𝑥 + 3) / \$ = ! 2. 𝑤 \$ !. 1 2 𝑑𝑤
11. 11. ! 10𝑥 . (2𝑥 + 3) # ! 𝑑𝑥 = ( ** ) 2𝑥. (2𝑥 + 3) . \$ − 2 7 (2𝑥 + 3) / \$ Jika kita substitusikan persamaan ( ** ) ke persamaan ( * ), maka diperoleh Di slide 8, kita mendapatkan persamaan ( * ), yaitu ∫ 15𝑥\$ . 2𝑥 + 3𝑑𝑥 = 5𝑥\$. (2𝑥 + 3) ! " − ∫ 10𝑥 . (2𝑥 + 3) ! "𝑑𝑥 ! 15𝑥\$ . 2𝑥 + 3𝑑𝑥 = 5𝑥\$. (2𝑥 + 3) % \$ − 2𝑥. (2𝑥 + 3) . \$ − 2 7 (2𝑥 + 3) / \$ = 5𝑥\$. (2𝑥 + 3) % \$ − 2𝑥. 2𝑥 + 3 . \$ + 2 7 (2𝑥 + 3) / \$
12. 12. LATIHAN ! 4𝑥 . 𝑠𝑒𝑐\$ 3𝑥 𝑑𝑥 ! 10𝑥\$ . sin 5𝑥 𝑑𝑥 1) 2)