Latihan 8.3 Thomas (Kalkulus Integral)
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This document contains 16 math problems involving integrals of trigonometric functions. The problems are worked out step-by-step showing the integration techniques used such as substitution, factorization of trigonometric expressions, and partial fraction decomposition. The solutions provide the integrated functions in terms of trigonometric functions and constants.
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Latihan 8.3 Thomas (Kalkulus Integral)
- 1. Nama : Nurkhalifah Anwar Kelas : A1 NIM : 1911041007 KALKULUS INTEGRAL LATIHAN 8.3 1. β« πππ 2π₯ ππ₯ Jawab : β« πππ 2π₯ ππ₯ = 1 2 π ππ 2π₯ + π 2. β« 3 π ππ π₯ 3 ππ₯ Jawab : β« 3 π ππ π₯ 3 ππ₯ = 3β« π ππ π₯ 3 ππ₯ = β9 πππ π₯ 3 + π 3. β« πππ 3 π₯ π ππ π₯ ππ₯ Jawab : β« πππ 3 π₯ π ππ π₯ ππ₯ = β« πππ 3 π₯ π(βπππ π₯) = β 1 4 πππ 4 π₯ + π 4. β« π ππ4 2π₯ πππ 2π₯ ππ₯ Jawab : β« π ππ4 2π₯ πππ 2π₯ ππ₯ = β« π ππ4 2π₯ π ( 1 2 π ππ 2π₯) = 1 10 π ππ5 2π₯ + π 5. β« π ππ3 π₯ ππ₯ Jawab : β« π ππ3 π₯ ππ₯ = β«π ππ2 π₯ β π ππ π₯ ππ₯ = β«(1 β πππ 2 π₯) π(βπππ π₯)
- 2. = 1 3 πππ 3 π₯ β πππ π₯ + π 6. β« πππ 3 4π₯ ππ₯ Jawab : β« πππ 3 4π₯ ππ₯ = β«πππ 2 4π₯ β πππ 4π₯ ππ₯ = β«(1 β π ππ2 4π₯) π ( 1 4 π ππ 4π₯) = 1 4 π ππ 4π₯ β 1 12 π ππ3 4π₯ + π 7. β« π ππ5 π₯ ππ₯ Jawab : β« π ππ5 π₯ ππ₯ = β« π ππ4 π₯ β π ππ π₯ ππ₯ = β«(1 β πππ 2 π₯)2 π(βπππ π₯) = β«(1 β 2πππ 2 π₯ + πππ 4 π₯) π(βπππ π₯) = βπππ π₯ + 2 3 πππ 3 π₯ β 1 5 πππ 5 π₯ + π 8. β« π ππ5 π₯ 2 ππ₯ Jawab : β« π ππ5 π₯ 2 ππ₯ = β«π ππ4 π₯ 2 β π ππ π₯ 2 ππ₯ = β«(1 β πππ 2 π₯ 2 ) 2 π (β2πππ π₯ 2 ) = β«(1 β 2πππ 2 π₯ 2 + πππ 4 π₯ 2 ) π (β2πππ π₯ 2 ) = β2πππ π₯ 2 + 4 3 πππ 3 π₯ 2 β 2 5 πππ 5 π₯ 2 + π 9. β« πππ 3 π₯ ππ₯ Jawab : β« πππ 3 π₯ ππ₯ = β« πππ 2 π₯ β πππ π₯ π(π ππ π₯)
- 3. = β«(1 β π ππ2 π₯)π(π ππ π₯) = π ππ π₯ β 1 3 π ππ3 π₯ + π 10. β« 3 πππ 5 3π₯ ππ₯ Jawab : β« 3 πππ 5 3π₯ππ₯ = 3 [β« πππ 4 3π₯ β πππ 3π₯ ππ₯] = 3 [β« (1 β π ππ2 3π₯)2 π ( 1 3 π ππ 3π₯)] = 3 [β« (1 β 2π ππ2 3π₯ + π ππ4 3π₯) π ( 1 3 π ππ 3π₯)] = [β« (1β 2π ππ2 3π₯ + π ππ4 3π₯) π(π ππ 3π₯)] = π ππ 3π₯ β 2 3 π ππ3 3π₯ + 1 5 π ππ5 3π₯ + π 11. β« πππ3 π₯ πππ 3 π₯ ππ₯ Jawab : β« π ππ3 π₯(1β π ππ2 π₯)πππ π₯ ππ₯ πππ ππ π’ = π ππ π₯ ππ’ = πππ π₯ ππ₯ ππ₯ = ππ’ πππ π₯ ππβπππππ π·πππππππβ: β« πππ3 π₯ πππ 3 π₯ ππ₯ = β« π’3 (1 β π’2 )ππ’ = β« π’3 β π’5 ππ’ = π’4 4 β π’6 6 + πΆ = 1 4 π ππ4 π₯ β 1 6 π ππ6 π₯ + πΆ 12. β«πππ 32π₯π ππ52π₯ ππ₯
- 4. Jawab : β«πππ 22π₯π ππ52π₯ πππ π₯ ππ₯ = β«(1 β π ππ22π₯) π ππ52π₯ πππ π₯ ππ₯ = β« (π ππ52π₯ β π ππ72π₯) πππ π₯ ππ₯ πππ ππ π’ = π ππ2π₯ ππ’ = 2 . πππ 2π₯ ππ₯ ππ’ 2 = πππ 2π₯ ππ₯ ππβπππππ ππππππππβ βΆ β« πππ 22π₯π ππ52π₯ πππ π₯ ππ₯ = β« (π’5 β π’7) ππ’ 2 = β« (π’5 β π’7) ππ’ 2 = 1 2 β« (π’5 β π’7) ππ’ = 1 2 ( 1 6 π’6 β 1 8 π’8) + πΆ = 1 12 π’6 β 1 16 π’8 + πΆ = 1 12 π ππ62π₯ β 1 16 π ππ82π₯ + πΆ 13. β«πππ 2π₯ ππ₯ = β― Jawab : β«πππ 2π₯ ππ₯ = β«(1 + πππ 2π₯ 2 ) ππ₯ = β« 1 2 ππ₯ + β« πππ 2π₯ 2 ππ₯ = 1 2 π₯ + 1 2 . 1 2 π ππ2π₯+ π = 1 2 π₯ + 1 4 π ππ2π₯ + π 14. β« π ππ2 π₯ ππ₯ Jawab :
- 5. β« πππ2 π₯ ππ₯ = β« (1βπππ 2π₯) 2 ππ₯ = β« 1 2 ππ₯ β β« πππ 2π₯ 2 ππ₯ = π₯ 2 β π ππ 2π₯ 2 Γ 1 2 + πΆ = π₯ 2 β π ππ 2π₯ 4 + πΆ 15. β« π ππ7 π¦ ππ¦ Jawab : β« π ππ7 π¦ ππ¦ = β« π ππ6 π₯.π ππ π₯ ππ₯ = β«(1 β πππ 2 π₯)3 .π ππ π₯ ππ₯ πππ ππ π’ = πππ π₯ ππ’ = βπ ππ π₯ ππ₯ ππ₯ = β ππ’ π ππ π₯ ππβπππππ ππππππππβ βΆ β« π ππ7 π¦ ππ¦ = β β«(1 β π’2 ) 3 ππ’ = ββ« (1 β3π’2 + 3π’4 β π’6) ππ’ = βπ’ + π’3 β 3π’5 5 + π’7 7 + π = βπππ π₯ + πππ 3π₯ β 3πππ 5π₯ 5 + πππ 7π₯ 7 + π 16. β« 7 πππ 7 π‘ ππ‘ Jawab : 7 β« πππ 7 π‘ ππ‘ = 7 β« πππ 6 π‘. πππ π‘ ππ‘ = β«(1 β π ππ2 π‘)3 . πππ π‘ ππ‘ πππ ππ π’ = π ππ π‘
- 6. ππ’ = πππ π‘ ππ‘ ππ‘ = ππ’ πππ π‘ ππβπππππ ππππππππβ βΆ β« 7 πππ 7 π‘ ππ‘ = 7 β«(1 β π’2 ) 3 ππ’ = 7 β«(1 β3π’2 + 3π’4 β π’6) ππ’ = 7π’ β 7π’3 + 21π’5 5 β 7 π’7 + π = 7π ππ π‘ β 7 π ππ3π‘ + 21π ππ5π‘ 5 β 7 π ππ7π‘+ π 17. β« 8 π ππ4 π₯ ππ₯ = Jawab : 8 β« π ππ4 π₯ ππ₯ = 8 β«(π ππ2 π₯)2 ππ₯ = 8 β« ( 1βπππ 2π₯ 2 ) 2 ππ₯ = 8 β« ( 1 4 β 2 πππ 2π₯ 4 + πππ 2 2π₯ 4 )ππ₯ = 8 β« 1 4 ππ₯ β 8β« 2 πππ 2π₯ 4 ππ₯ + 8 β« πππ 2 2π₯ 4 ππ₯ = 8 4 β« ππ₯ β 16 4 β« πππ 2π₯ ππ₯ + 8 4 β« πππ 2 2π₯ ππ₯ = 2 β« ππ₯ β 4 β« πππ 2π₯ ππ₯ + 2 β« ( 1+πππ 4π₯ 2 )ππ₯ = 2π₯ β 4 ( 1 2 )π ππ 2π₯ + π₯ + ( 1 4 )π ππ 4π₯ + π = 3π₯ β 2 π ππ 2π₯ + ( 1 4 )π ππ 4π₯ + π 18. β« 8 πππ 4 π₯ 2ππ₯ ππ₯ Jawab : Misal : π’ = 2ππ₯ cos2 π’ = 1 + cos2π’ 2 cos2 2π’ = 1 + cos4π’ 2
- 7. ππ’ = 2π ππ₯ ππ’ 2π = ππ₯ Sehingga, β« 8 πππ 4 π₯ 2ππ₯ ππ₯ = β« 8πππ 4 π’ ππ’ 2π = 8 2π β« πππ 4 π’ ππ’ = 4 π β«(πππ 2 π’)2 ππ’ = 4 π β« ( 1+πππ 2π’ 2 ) 2 ππ’ = 4 π β« 1 4 (1 + πππ 2π’)2 ππ’ = 1 π β«(1 + 2 πππ 2π’ + πππ 2 2π’) ππ’ = 1 π β«(1 + 2 πππ 2π’ + ( 1+πππ 4π’ 2 )) ππ’ = 1 π β« (1 + 2 πππ 2π’ + 1 2 (1 + πππ 4π’))ππ’ = 1 π (π’ + π ππ 2π’ + 1 2 π’ + 1 8 π ππ 4π’) + πΆ = 2π₯ + 1 π π ππ 4ππ₯ + π₯ + 1 8 π ππ 8ππ₯ + πΆ 19. β« 16π ππ2 π₯πππ 2 π₯ ππ₯ Jawab : β« 16π ππ2 π₯πππ 2 π₯ ππ₯ = β« 4 Γ 4π ππ2 π₯πππ 2 π₯ ππ₯ = β« 4 Γ 22 π ππ2 π₯πππ 2 π₯ ππ₯ = β« 4(2π ππ π₯ πππ π₯)2 ππ₯ = 4 β« π ππ2 2π₯ ππ₯ Misal: π’ = 2π₯ ππ’ = 2 ππ₯
- 8. ππ₯ = 1 2 ππ’ ππβπππππ ππππππππβ βΆ β« 16π ππ2 π₯πππ 2 π₯ ππ₯ = 4 β« π ππ2 π’ ( 1 2 )ππ’ = 4 ( 1 2 )β« π ππ2 π’ ππ’ = 2 β« ( 1βπππ 2π’ 2 )ππ’ = β«(1 β πππ 2π’)ππ’ = β« ππ’ β β« πππ 2π’ ππ’ = π’ β 1 2 π ππ 2π’ + π = 2π₯ β 1 2 π ππ 4π₯ + π 20. β« 8 π ππ4 π¦ πππ 2 π¦ ππ¦ Jawab : 8 β«π ππ4 π¦ πππ 2 π¦ ππ¦ = 8β«(π ππ2 π¦)2 πππ 2 π¦ ππ¦ = 8β« ( 1βπππ 2π¦ 2 ) 2 ( 1βπππ 2π¦ 2 ) ππ¦ = 8β« 1 4 (1 β 2πππ 2π¦ + πππ 2 2π¦) 1 2 (1 + πππ 2π¦) = β«(1 + πππ 2π¦ β 2πππ 3π¦ β 2πππ 2 2π¦ + πππ 2 2π¦ + πππ 3 2π¦) ππ¦ = β«(1 β πππ 2π¦ β πππ 2 2π¦ + πππ 3 2π¦) ππ¦ = β«(1 β πππ 2π¦ β ( 1 2 + 1 2 πππ 4π¦) + (πππ 2 2π¦)πππ 2π¦) ππ¦ = β«( 1 β πππ 2π¦ β 1 2 β 1 2 πππ 4π¦ + (1 β π ππ2 2π¦) πππ 2π¦ ππ¦
- 9. = β«( 1 2 β πππ 2π¦ β 1 2 πππ 4π¦ + πππ 2π¦ β π ππ2 2π¦ πππ 2π¦) ππ¦ = β«( 1 2 β 1 2 πππ 4π¦)ππ¦ β β«(π ππ2 2π¦) π(π ππ π¦) = 1 2 π¦ β 1 8 π ππ 4π¦ β 1 6 π ππ3 2π¦ + πΆ 21. β« 8 πππ 3 2π π ππ 2π ππ Penyelesaian Misal u = πππ 2π du = β π ππ 2π π(2π) d (2π) = β ππ’ π ππ 2π Sehingga β« 8 πππ 3 2π π ππ 2π ππ = 1 2 β« 8 πππ 3 2π π ππ 2π π (2π) = 1 2 β« 8 π’3 π ππ 2π . β ππ’ π ππ 2π = β 1 2 ( 8 4 π’4 ) + πΆ = β πππ 4 2π + πΆ 22. β« π ππ2 2π πππ 3 2π ππ π 2 0 = 1 2 β« π ππ2 2π πππ 3 2π π(2π) π 2 0 = 1 2 β« π ππ2 2π πππ 2 2ππππ 2π π(2π) π 2 0 = 1 2 β« [π ππ2 2π (1β π ππ2 2π)] π(π ππ 2π) π 2 0 = 1 2 β« (π ππ2 2π β π ππ4 2π) π(π ππ 2π) π 2 0 = 1 2 ( 1 3 π ππ3 2π β 1 5 π ππ5 2π)] 0 π 2 β = 1 6 π ππ3 2π β 1 10 π ππ5 ]0 π 2 β ] =[ 1 6 π ππ3 2( π 2 ) β 1 10 π ππ5 2( π 2 )] β [ 1 6 π ππ3 2(0)β 1 10 π ππ5 2(0)]
- 10. = 0 23. β« β 1βπππ π₯ 2 ππ₯ 2π 0 = β« βπ ππ2 1 2 π₯ ππ₯ 2π 0 = β« π ππ 1 2 π₯ ππ₯ 2π 0 = 2 β« π ππ 1 2 π₯ π ( 1 2 π₯) 2π 0 = Λ2 πππ 1 2 π₯ + πΆ]0 2π =[β2πππ 1 2 (2π)] β [β2 πππ 1 2 (0)] = β2πππ 1 2 (2π) + 2 πππ 1 2 (2π) = 2 + 2 = 4 24. β« β1 β πππ 2π₯ ππ₯ π 0 = β« βπ ππ2π₯ + πππ 2π₯ β πππ 2π₯ + π ππ2 π₯ π 0 ππ₯ = β« βπ ππ2π₯ + π ππ2π₯ π 0 ππ₯ = β« β2 .π ππ2 π₯ ππ₯ π 0 = β« β2 .π ππ π₯ π 0 = β β2 πππ π₯]0 π = [β β2πππ (π)] β [ββ2πππ (0)] = β β2πππ (π) + β2πππ (0) = β β2 .β1 + β2 .1 =β2 + β2 = 2β2 25. β« β1 β π ππ2π‘ ππ‘ π 0 = β« βπ ππ2π‘ + πππ 2π‘ β π ππ2π‘ ππ‘ π 0 = β« βπππ 2π‘ ππ‘ π 0 = β« πππ π‘ ππ‘ π 0 = π ππ π‘ + πΆ]0 π = π ππ π β π ππ 0 = 0 Λ 0
- 11. = 0 26. β« β1 β πππ 2π ππ π 0 = β« βπ ππ2π + πππ 2π β πππ 2π ππ π 0 = β« βπ ππ2π ππ π 0 = βπππ π]2 π = βπππ π + πππ 0 = β(β1)+ 1 = 2 27. β« π ππ2 π₯ β1βπππ π₯ π 3 β π 2 β ππ₯ = β« π ππ2 π₯ β1βπππ π₯ π 3 β π 2 β . β1+πππ π₯ β1+πππ π₯ ππ₯ = β« π ππ2 π₯ β1+πππ π₯ β1βπππ π₯ π 3 β π 2 β ππ₯ = β« π ππ2 π₯ β1+πππ π₯ π ππ π₯ π 3 β π 2 β ππ₯ = β« π ππ π₯β1 + πππ π₯ π 3 β π 2 β ππ₯ = β« π ππ π₯ π’ 1 2 β π 3 β π 2 β . β ππ’ π ππ π₯ = ββ« π’ 1 2 β ππ’ π 3 β π 2 β = β( 2 3 π’ 3 2 β )]π 3 π 2 = β 2 3 (1 + πππ π₯) 3 2 β ]π 3 π 2 = β 2 3 (1 + πππ ( π 2 )) 3 2 β + 2 3 (1 + πππ ( π 3 )) 3 2 β = β 2 3 + 2 3 ( 3 2 ) 3 2 β = β 3 2 β 2 3 28. β« β1 + π ππ π₯ ππ₯ π 6 β 0 = β« β1+ π ππ π₯ . β1βπ ππ π₯ β1βπ ππ π₯ ππ₯ π 6 β 0 = β« β1βπ ππ2 π₯ β1βπ ππ π₯ ππ₯ π 6 β 0 = β« πππ π₯ β1βπ ππ π₯ ππ₯ π 6 β 0 Misal u = 1 + cosπ₯ du = β sin π₯ ππ₯ dx = β ππ’ sinπ₯ Misal u = 1 β sin π₯ du = β cosπ₯ ππ₯ dx = β ππ’ cosπ₯
- 12. = β« πππ π₯ .π’ β 1 2 . β ππ’ πππ π₯ π 6 β 0 = ββ« π’ β 1 2 π 6 β 0 ππ’ = β2 π’ 1 2 β + πΆ]0 π 6 = β2 β1β π ππ π₯ + πΆ]0 π 6 = β2 β1 β π ππ π 6 + 2β1 β π ππ 0 = β2β1 β 1 2 + 2 β1β 0 = β2 β 1 2 + 2 29. β« πππ 4 π₯ β1βπ ππ π₯ ππ₯ π 5π 6 β = β« πππ 4 π₯ β1βπ ππ π₯ . β1+π ππ π₯ β1+π ππ π₯ π 5π 6 β = β« πππ 4 π₯β1+π ππ π₯ β1βπ ππ2 π₯ π 5π 6 β = β« πππ 4 π₯ β1+π ππ π₯ βπππ 2 π₯ π 5π 6 β = β« πππ 4 π₯ β1+π ππ π₯ πππ π₯ π 5π 6 β = β« πππ 4 π₯ β1+π ππ π₯ β πππ π₯ π 5π 6 β = β« βπππ 3 π₯β1 + π ππ π₯ π 5π 6 β =β« βπππ π₯ (1β π ππ2 π₯)β1 + π ππ π₯ π 5π 6 β = ββ« (1 β π ππ2 π₯)β1+ π ππ π₯ πππ π₯ ππ₯ π 5π 6 β = ββ« [1 β (π’ β 1)2]βπ’ 1 3 2 β ππ’ = ββ« [1 β (π’2 β 2π’ + 1)]βπ’ 1 3 2 β ππ’ =ββ« (βπ’ 5 2 β + 2π’ 3 2 β ) 1 3 2 β ππ’ =β« (π’ 5 2 β + 2π’ 3 2 β ) 1 3 2 β du Misal u = 1 + sin π₯ du = cosπ₯ ππ₯ xβ 5π 6 β π’ = 3 2 xβ π β π’ = 1
- 13. = 2 7 π’ 7 2 β β 4 5 π’ 5 2 β ] 3 2 β 1 =( 2 7 β 4 5 ) β [ 2 7 ( 3 2 ) 7 2 β β 4 5 ( 3 2 ) 5 2 β ] = β 18 35 β [ 2 7 . 3 7 2 β 2 7 2 β β 23 5 . 3 5 2 β 2 5 2 β ] = β 18 35 β 3 5 2 β 2 1 2 β [ 3 7 . 4 β 1 5 ] = β 18 35 β 9β6 2 [β 13 140 ] = 117β6 280 β 18 35 30. β« β1 β π ππ 2π₯ ππ₯ 3π 4 β π 2 β = β« β1 β π ππ 2π₯ . β1+π ππ 2π₯ β1+π ππ 2π₯ ππ₯ 3π 4 β π 2 β = β« β1βπ ππ2 2π₯ β1+π ππ 2π₯ ππ₯ 3π 4 β π 2 β = β« βπππ 2 π₯ β1+π ππ 2π₯ ππ₯ 3π 4 β π 2 β = ββ1+ π ππ 2π₯]π 2 3π 4 = ββ1 + π ππ 2( 3π 4 ) + β1 + π ππ 2( π 2 ) = ββ1 β 1 + 1 = 1 31.β«π₯β1 β πππ 2π₯ ππ₯ β«π₯βπ ππ2π₯+ πππ 2π₯ β (πππ 2π₯ β π ππ2π₯) ππ₯ β«π₯ β2π ππ2π₯ ππ₯ β«π₯. β2 π πππ₯ ππ₯ β2 (π ππ π₯ β π₯ πππ π₯) + π 32. β«β(1 β πππ 2π‘) 3 2ππ‘ β«(π ππ2π‘) 3 2 ππ‘
- 14. β«π ππ3 π‘ ππ‘ β«(1 β πππ 2 π‘) π(βπππ π‘) β πππ π‘ + 1 3 πππ 3 π‘+ π 33. β«π ππ2π₯π‘ππ π₯ ππ₯ β«π‘ππ π₯ π(π‘ππ π₯) 1 2 π‘ππ2π₯ + π 34. β« π ππ π₯ π‘ππ2 π₯ ππ₯ β« π ππ π₯ π‘ππ2 π₯ ππ₯ = β« π ππ π₯ (π ππ2 π₯ β 1)ππ₯ = β« π ππ3 π₯ β π ππ π₯ ππ₯ = β« π ππ2 π₯ π ππ π₯ β π ππ π₯ ππ₯ = β«(1 + π‘ππ2 π₯ )π ππ π₯ β π ππ π₯ ππ₯ = β« π ππ π₯ + π‘ππ2 π₯ π πππ₯ β π ππ π₯ dx = β« π‘ππ2 π₯ π πππ₯ = π ππ π₯ + πΆ 35. β« π ππ3 π₯ π‘ππ π₯ ππ₯ β« π ππ3 π₯ π‘ππ π₯ ππ₯ = β« π ππ2 π₯ π ππ π₯ π‘ππ π₯ ππ₯ = β« π ππ2 π₯ π (π ππ π₯) = 1 3 π ππ3 π₯ + πΆ 36. β« π ππ3 π₯ π‘ππ2 π₯ ππ₯ β« π ππ3 π₯ π‘ππ2 π₯ ππ₯ = β« π ππ2 π₯ π‘ππ2 π₯ π ππ π₯ π‘ππ π₯ ππ₯ = β« π ππ2 π₯ (π ππ2 π₯ β 1) π(π ππ π₯) = β« π ππ4 π₯ β π ππ2 π₯ π(π ππ π₯) = π ππ5 π₯ 5 β π ππ3 π₯ 3 + πΆ
- 15. 37. β« π ππ2 π₯ π‘ππ2 π₯ ππ₯ β« π ππ2 π₯ π‘ππ2 π₯ ππ₯ = β« π‘ππ2 π₯ π(π‘πππ₯) = π‘ππ3 π₯ 3 + πΆ 38. β« π ππ4 π₯ π‘ππ2 π₯ ππ₯ = β« π ππ2 π₯ π ππ2 π₯ π‘ππ2 π₯ ππ₯ =β«(1 + π‘ππ2 π₯) π‘ππ2 π₯ π (π‘ππ π₯) =β« π‘ππ2 π₯ + π‘ππ4 π₯ π(π‘ππ π₯) = 1 3 π‘ππ3 π₯ + 1 5 π‘ππ5 + πΆ 39. β« 2 π ππ3 π₯ ππ₯ Menggunakan rumus : β« π πππ ππ₯ ππ₯ = π πππ β2 ππ₯ π‘ππ ππ₯ π(πβ1) + πβ2 πβ1 β« π πππβ2 ππ₯ ππ₯ = 2 β« π ππ3 π₯ ππ₯ = 2 π ππ3β2 π₯ π‘ππ π₯ (3β1) + 3β2 3β1 β« π ππ3β2 π₯ ππ₯ = 2 π ππ π₯ π‘ππ π₯ 2 + 1 2 β« π ππ π₯ ππ₯ =π ππ π₯ π‘ππ π₯ + 1 2 β« π ππ π₯ ππ₯ = π ππ π₯ π‘ππ π₯ + 1 2 ππ|π ππ π₯ + π‘ππ π₯| + π 40. β« ππ₯ π ππ3 ππ₯ ππ₯ = ππ₯ β« π ππ3 ππ₯ ππ₯ = ππ₯ π ππ3β2 ππ₯ π‘ππ ππ₯ ππ₯(3β1) + 3β2 3β1 β« π ππ3β2 ππ₯ ππ₯ = π ππ ππ₯ π‘ππ ππ₯ 2 + 1 2 β« π ππ ππ₯ ππ₯ = 1 2 π ππ ππ₯ + 1 2 β« π ππ ππ₯ ππ₯ = 1 4 π ππ ππ₯ + ππ|π ππ ππ₯ + π‘ππ ππ₯| + πΆ 41. β« π ππ4 πππ
- 16. = β«(1 + π‘ππ2 π)π ππ2 πππ = β« π ππ2 πππ + β« π‘ππ2 ππ ππ2 πππ = = π‘πππ + 1 3 π‘ππ3 π + πΆ = π‘πππ + 1 3 π‘πππ(π ππ2 π β 1) + πΆ = = 1 3 π‘ππππ ππ2 π + 2 3 π‘πππ + πΆ . (π’ = π‘πππ₯) β¦(π‘πππ₯)β² = π ππ2 π₯ . β« π’π ππ’ = π’π+1 π+1 + π . 1 3 π‘ππππ ππ2 π + 2 3 π‘πππ + πΆ 42. β« 3π ππ4 3π₯ ππ₯ = Jawab : β« 3π ππ4 3π₯ ππ₯ = 3 β«(π ππ2 3π₯)(π ππ2 3π₯)ππ₯ = 3β«(π‘ππ2 3π₯ + 1)(π ππ2 3π₯)ππ₯ = 3β«(π‘ππ2 3π₯ + 1)π 1 3 π‘ππ 3π₯ = 3 β 1 3 [β« π‘ππ2 3π₯π π‘ππ π₯ + β« 1 π π‘ππ 3π₯] = 1 β [ 1 3 π‘ππ3 3π₯ + π‘ππ 3π₯] + πΆ = 1 3 π‘ππ3 3π₯ + π‘ππ 3π₯ + πΆ 43. β« ππ π4 π ππ π/2 π/4 = Jawab : β« ππ π4 π ππ π/2 π/4 = β«(1 + πππ‘2 π)ππ π2 π ππ = β«((1 + πππ‘2 π)π (βcotπ) = β« 1 π(βcot π) β β« πππ‘2 π π (cotπ)
- 17. = β cotπ β 1 3 πππ‘3 π + πΆ 44. β« π ππ6 π₯ ππ₯ = β« π ππ4 π₯ β π ππ2 π₯ ππ₯ = β«(π‘ππ2 π₯ + 1)2 π ππ2 π₯ ππ₯ = β«(π‘ππ4 π₯ + 2π‘ππ2 π₯ + 1)π ππ2 π₯ ππ₯ = β« π‘ππ4 π₯ π ππ2 π₯ ππ₯ + 2 β« π‘ππ2 π₯ π ππ2 π₯ ππ₯ + β« π ππ2 π₯ ππ₯ = 1 5 π‘ππ5 π₯ + 2 3 π‘ππ3 π₯ + π‘ππ π₯ + πΆ 45. β« 4 π‘ππ3 π₯ ππ₯ = Jawab : β« 4π‘ππ3 π₯ ππ₯ = 4 β« π‘ππ2 π₯ π‘ππ π₯ ππ₯ = 4β«(π ππ2 π₯ β 1)π‘ππ π₯ ππ₯ = 4β«(π‘ππ π₯ π ππ2 π₯ β π‘ππ π₯) ππ₯ = 4β« π‘ππ π₯ π (π‘ππ π₯) β β« π‘ππ π₯ ππ₯ = 4 ( 1 2 π‘ππ2 π₯ β ππ|π ππ π₯| + πΆ = 2 π‘ππ2 π₯ β 4ππ|π ππ π₯| + πΆ 46. β« 6 π‘ππ4 π₯ ππ₯ π 4 β π 4 = Jawab : β« 6 π‘ππ4 π₯ ππ₯ π 4 β π 4 = 6β«(π‘ππ2 π₯)(π‘ππ2 π₯) ππ₯ = 6 β«(π ππ2 β 1)π‘ππ2 π₯ ππ₯ = 6 β«(π‘ππ2 π₯π ππ2 π₯ β π‘ππ2 π₯)ππ₯ = 6 β« π‘ππ2 π₯ π(π‘πππ₯) β β«(π ππ2π₯ β 1)ππ₯ = 6 β« π‘ππ2 π₯ π(π‘πππ₯) β β« π ππ2 π₯ ππ₯ + β« 1 ππ₯
- 18. = 6 ( 1 3 π‘ππ3 π₯ β π‘ππ π₯ + π₯) + πΆ = 2π‘ππ3 π₯ β 6 π‘ππ π₯ + 6π₯ + πΆ 47. β« π‘ππ5 π₯ ππ₯ = Jawab : β« π‘ππ5 π₯ ππ₯ = β«(π‘ππ2 π₯)2 π‘ππ π₯ ππ₯ = β«(π ππ2 π₯ β 1)2 π‘ππ π₯ ππ₯ = β«(π ππ4 π₯ β 2π ππ2 π₯ + 1)π‘ππ π₯ ππ₯ = β«(π‘ππ π₯ π ππ4 π₯ β 2π‘ππ π₯ π ππ2 π₯ + π‘ππ π₯) ππ₯ = β«(π‘ππ π₯ (1+ π‘ππ2 π₯)π ππ2 π₯ β 2 π‘ππ π₯ π ππ2 π₯ + π‘ππ π₯) ππ₯ = β«(π‘ππ π₯ + π‘ππ3 π₯) π ππ2 π₯ ππ₯ β 2 β« π‘ππ π₯ π ππ2 π₯ ππ₯ + β« π‘ππ π₯ ππ₯ = β«(π‘ππ π₯ + π‘ππ3 π₯) π π‘ππ π₯ β 2 β« π‘ππ π₯ π π‘ππ π₯ + β« π‘ππ π₯ ππ₯ = β« π‘ππ π₯ π π‘ππ π₯ + β« π‘ππ3 π₯ π π‘ππ π₯ β 2 β« π‘ππ π₯ π π‘ππ π₯ + β« π‘ππ π₯ ππ₯ = 1 2 π‘ππ2 π₯ + 1 4 π‘ππ4 π₯ β 2 2 π‘ππ2 π₯ + ππ|π ππ π₯| + πΆ = 1 4 π‘ππ4 π₯ β 1 2 π‘ππ2 π₯ + ππ|π ππ π₯| + πΆ 48. β« πππ‘6 2π₯ ππ₯ = β«(ππ π2 2π₯ β 1)3 ππ₯ = β«(ππ π6 2π₯ β 3 ππ π4 2π₯ + 3 ππ π2 2π₯ + 1) ππ₯ = β«[(ππ π4 2π₯ β 3 ππ π2 2π₯ + 3)(ππ π2 2π₯) + 1] ππ₯ = β«[((1 + πππ‘2 2π₯)2 β 3(1 + πππ‘2 2π₯) + 3)(ππ π2 2π₯) + 1] ππ₯
- 19. = β«[(1 + 2πππ‘2 2π₯ + πππ‘4 2π₯ β 3 β 3πππ‘2 2π₯ + 3)(ππ π2 2π₯) + 1] ππ₯ = β« [(1 + πππ‘4 2π₯ β πππ‘2 2π₯) π (β 1 2 πππ‘ 2π₯)] + β«1 ππ₯ = [ β 1 2 (β« 1 ππππ‘ 2π₯ + β«πππ‘4 2π₯ π πππ‘ 2π₯ β β«πππ‘2 2π₯ π πππ‘ 2π₯)] + β«1 ππ₯ = β 1 2 (πππ‘ 2π₯ + 1 5 πππ‘5 2π₯ β 1 3 πππ‘3 2π₯) + π₯ + π = β 1 2 πππ‘ 2π₯ β 1 10 πππ‘5 2π₯ + 1 6 πππ‘3 2π₯ + π₯ + π = β 1 10 πππ‘5 2π₯ + 1 6 πππ‘3 2π₯ β 1 2 πππ‘ 2π₯ + π₯ + π 49. β« πππ‘3 π₯ππ₯ =β«(πππ‘2 π₯)(πππ‘π₯)ππ₯ = β«(ππ π2 π₯ β 1)(πππ‘π₯)ππ₯ = β«(πππ‘π₯ππ π2 π₯ β πππ‘π₯)ππ₯ = β« πππ‘π₯ππ π2 π₯ ππ₯ β πππ‘ π₯ ππ₯ Misalkan u = cot x du = βππ π2 π₯ ππ₯ = β« βπ’ ππ’ β β« πππ‘ π₯ = β 1 2 π’2 β ππ|π πππ₯|+C 50. β« 8 πππ‘4 π₯ ππ₯ =8β« πππ‘4 π₯ ππ₯ =8 β« πππ‘2 π₯πππ‘2 π₯ ππ₯ =8β« πππ‘2 π₯ (ππ π2 π₯ β 1)ππ₯ =8 β« πππ‘2 π₯ππ π2 π₯ β πππ‘2 π₯ ππ₯
- 20. =8 β« πππ‘2 π₯ππ π2 π₯ ππ₯ β β« πππ‘2 π₯ ππ₯ Misalkan = u = cot x du = βππ π2 π₯ ππ₯ =-8β« π’2 ππ’ β β« πππ‘2 π₯ ππ₯ =-8β« π’2 ππ’ β β« ππ π2 π₯ β 1 ππ₯ =-8 β« π’2 ππ’ β β« ππ π2 ππ₯ β β« 1 ππ₯ = β 8 3 πππ‘3 π₯ + 8 πππ‘ π₯ + 8 π₯ + πΆ