Soal integral
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Dipi76
New member
Bls: Soal integral
Pake tulisan biasa aja ya, mas...
)
Aku rada gaptek kalo harus pake fitur penulisan yang sudah disediakan itu...
1. sin^4(x)
= [sin^2(x)]^2
= [(1 -cos(2x)/2)]^2
= 1/4[1 + cos^2(2x) - 2cos(2x)]
= 1/4[1 - 2cos(2x) + (1 + cos(4x)/2 ]
= (1/4) - (1/2)cos(2x) + (1/8) + (1/8)cos(4x)
= (3/8) - (1/2)cos(2x) + (1/8)cos(4x)
maka ∫sin^4(x)dx = 3/8∫dx - 1/2∫cos(2x) dx + 1/8∫cos(4x) dx
∫sin^4(x)dx = (3/8)x - (1/2)sin(2x)(1/2) + (1/8)sin(4x)(1/4) + C
∫sin^4(x)dx = (3/8)x - (1/4)sin(2x) + (1/32)sin(4x) + C
2. ∫ cos^4(x) dx = ∫ (cos^2x)^2 dx
karena cos^2x = (1/2)[1 + cos(2x)], maka menjadi
∫ {(1/2)[1 + cos(2x)]}^2 dx = ∫ (1/4)[1 + cos(2x)]^2 dx
∫ (1/4)[1 + 2cos(2x) + cos^2(2x)] dx
= (1/4) ∫ dx + (1/4) 2 ∫ cos(2x) + (1/4) ∫ cos^2(2x) dx
= (1/4) ∫ dx + (1/2) ∫ cos(2x) + (1/4) ∫ cos^2(2x) dx
= (1/4)x + (1/2) (1/2)sin(2x) + (1/4) ∫ cos^2(2x) dx
= (1/4)x + (1/4)sin(2x) + (1/4) ∫ cos^2(2x) dx
dari hasil di atas, maka
cos^2(2x) = (1/2){1 + cos[2(2x)]} = (1/2)[1 + cos(4x)]
oleh karena itu,
(1/4)x + (1/4)sin(2x) + (1/4) ∫ (1/2)[1 + cos(4x)] dx
= (1/4)x + (1/4)sin(2x) + (1/4)(1/2) ∫ [1 + cos(4x)] dx
= (1/4)x + (1/4)sin(2x) + (1/8) ∫ dx + (1/8) ∫ cos(4x) dx
= (1/4)x + (1/4)sin(2x) + (1/8)x + (1/8) (1/4)sin(4x) + C
= [(1 + 2)/8]x + (1/4)sin(2x) + (1/32)sin(4x) + C
Maka,
∫ cos^4(x) dx = (3/8)x + (1/4)sin(2x) + (1/32)sin(4x) + C
-dipi-
Pake tulisan biasa aja ya, mas...
)Aku rada gaptek kalo harus pake fitur penulisan yang sudah disediakan itu...
1. sin^4(x)
= [sin^2(x)]^2
= [(1 -cos(2x)/2)]^2
= 1/4[1 + cos^2(2x) - 2cos(2x)]
= 1/4[1 - 2cos(2x) + (1 + cos(4x)/2 ]
= (1/4) - (1/2)cos(2x) + (1/8) + (1/8)cos(4x)
= (3/8) - (1/2)cos(2x) + (1/8)cos(4x)
maka ∫sin^4(x)dx = 3/8∫dx - 1/2∫cos(2x) dx + 1/8∫cos(4x) dx
∫sin^4(x)dx = (3/8)x - (1/2)sin(2x)(1/2) + (1/8)sin(4x)(1/4) + C
∫sin^4(x)dx = (3/8)x - (1/4)sin(2x) + (1/32)sin(4x) + C
2. ∫ cos^4(x) dx = ∫ (cos^2x)^2 dx
karena cos^2x = (1/2)[1 + cos(2x)], maka menjadi
∫ {(1/2)[1 + cos(2x)]}^2 dx = ∫ (1/4)[1 + cos(2x)]^2 dx
∫ (1/4)[1 + 2cos(2x) + cos^2(2x)] dx
= (1/4) ∫ dx + (1/4) 2 ∫ cos(2x) + (1/4) ∫ cos^2(2x) dx
= (1/4) ∫ dx + (1/2) ∫ cos(2x) + (1/4) ∫ cos^2(2x) dx
= (1/4)x + (1/2) (1/2)sin(2x) + (1/4) ∫ cos^2(2x) dx
= (1/4)x + (1/4)sin(2x) + (1/4) ∫ cos^2(2x) dx
dari hasil di atas, maka
cos^2(2x) = (1/2){1 + cos[2(2x)]} = (1/2)[1 + cos(4x)]
oleh karena itu,
(1/4)x + (1/4)sin(2x) + (1/4) ∫ (1/2)[1 + cos(4x)] dx
= (1/4)x + (1/4)sin(2x) + (1/4)(1/2) ∫ [1 + cos(4x)] dx
= (1/4)x + (1/4)sin(2x) + (1/8) ∫ dx + (1/8) ∫ cos(4x) dx
= (1/4)x + (1/4)sin(2x) + (1/8)x + (1/8) (1/4)sin(4x) + C
= [(1 + 2)/8]x + (1/4)sin(2x) + (1/32)sin(4x) + C
Maka,
∫ cos^4(x) dx = (3/8)x + (1/4)sin(2x) + (1/32)sin(4x) + C
-dipi-
hamtaro_87
New member
Bls: Soal integral
Jadi kangen masa sma dl,
Krn guru Mat gw cinta bgt ma 'Integral' khususny 'Integral Parsial'..
Dohh, tiap hr pr Integral itu berlembar2 boo...
^^
Jadi kangen masa sma dl,
Krn guru Mat gw cinta bgt ma 'Integral' khususny 'Integral Parsial'..
Dohh, tiap hr pr Integral itu berlembar2 boo...
^^
tiaseptiani
New member
waduh ga ngerti