Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Tentukanlah Integral dari :
∫ sin^4 x cos^3 x dx
Jawab:
Misalkan: u = sin x
Maka: du/dx = cosx
Sehingga: dx = du/cosx
∫ sin^4 x cos^3 x dx = ∫u^4 cos^3 x du/cosx
∫ sin^4 x cos^3 x dx = ∫ u^4 cos^3 x (du/cos x)
∫ sin^4 x cos^3 x dx = ∫ u^4 cos^2 x du
Karena: cos^2 x = 1 – sin^2 x
Maka:
∫ sin^4 x cos^3 x dx = ∫ u^4 (1- sin^2 x) du
Karena: u = sin x
Maka:
∫ sin^4 x cos^3 x dx = ∫ u^4 (1- u^2) du
∫ sin^4 x cos^3 x dx = ∫ (u^4- u^6) du
∫ sin^4 x cos^3 x dx = 1 /(4+1) u^(4+1) – 1 /(6+1) u^(6+1) + C
∫ sin^4 x cos^3 x dx = 1 /5 u^5 – 1 /7 u^7 + C
∫ sin^4 x cos^3 x dx = 1 /5 sin^5 x – 1 /7 sin^7 x + C
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Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Tentukanlah integral dari : ∫ sin^4 x dx
Jawab:
∫ sin ^n u du = – (sin^(n-1) u cosu) /n + (n-1) /n ∫ sin^(n-2) u du + c
Maka:
∫ sin^4 x dx = – (sin^(4-1) x cosx) /4 + (4-1)/4 ∫ sin^(4-2) x dx
∫ sin^4 x dx = – (sin^3 x cosx) /4 + 3/4 ∫ sin^2 x dx
∫ sin^4 x dx = – (sin^3 x cosx) /4 + 3/4 [- (sin^(2-1) x cosx) /2 + (2-1)/2 ∫ sin^(2-2) x dx ]
∫ sin^4 x dx = – (sin^3 x cosx) /4 + 3/4 [- (sin^1 x cosx) /2 + 1/2 ∫ sin^0 x dx ]
∫ sin^4 x dx = – (sin^3 x cosx) /4 + 3/4 [- (sin x cosx)/2 + 1/2 ∫ dx ]
∫ sin^4 x dx = – ( sin^3 x cosx) /4 + 3/4 [- (sin x cosx)/2+1/2 (x) ] + C
∫ sin^4 x dx = – (sin^3 x cosx) /4 – (3 sin x cosx) / 8 + (3 x) / 8 + C
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Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Tentukanlah integral dari : ∫ cos^4 x dx
Jawab:
∫cos ^n u du = (cos^(n-1) u sin u) / n + (n-1) /n ∫ cos^(n-2) u du + c
Maka:
∫ cos^4 x dx = (cos^3 x sin x) /4 + (4-1) /4 ∫ cos^2 x dx
∫ cos^4 x dx = (cos^3 x sin x) /4 + 3/4 ( ( cos x sin x) /2 + (2-1)/2 ∫ (cosx)^(2-2) dx )
∫ cos^4 x dx = (cos^3 x sin x) /4 + 3/4 ( ( cos x sin x) /2 + 1/2 ∫ (cosx)^0 dx )
∫ cos^4 x dx = (cos^3 x sin x) /4 + 3/4 ( ( cos x sin x) /2 + 1/2 ∫ dx )
∫ cos^4 x dx = (cos^3 x sin x) /4 + 3/4 (( cos x sin x) /2 + 1/2 (x) ) + C
∫ cos^4 x dx = (cos^3 x sin x) /4 + ( 3 cos x sin x) /8 + 3/8 x + C
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Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Tentukanlah Integral dari :
-
∫x e^( x^2 ) dx
-
∫ dx/(x lnx )
-
∫ (e^x dx)/(1+2e^x )
Jawab:
1. Misalkan: u = x^2
Maka: du/dx = 2x
Sehingga : dx = du/2x
∫ x e^( x^2 ) dx = ∫ x e^u du/2x = 1/2 ∫ e^u du = 1/2 e^u + C = 1/2 e^(x^2 ) + C
2. Misalkan: u = ln x
Maka: du/dx = 1/x
Sehingga : dx = x du
∫ dx /(x lnx ) = ∫ (x du) /(x u) = ∫ ( du) / u = ln|u| + C = ln |lnx | + C = ln ln x + C
3. Misalkan: u = 1 + 2e^x
Maka: du/dx = 2e^x
Sehingga : dx = du /2e^x
∫ (e^x dx) /(1+2e^x ) = ∫ (e^x ) /u (du /2e^x) = 1/2 ∫ du /u = 1/2 ln|u| + C = 1/2 ln|1 + 2e^x| + C
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Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Tentukanlah Integral dari :
-
∫ (x dx) /(x+1)
-
∫ e^sinx cos x dx
Jawab:
1. Misalkan: u = x + 1
maka: x = u – 1
Maka : du/dx=1
Sehingga: dx = du
∫ (x dx) /(x+1) = ∫ (u-1) /u du = ∫ u /u du – ∫ du /u = ∫ du – ∫ du /u = x – ln |u| + C = x – ln |x+1| + C
2. Misalkan: u = sin x
Maka : du/dx = cosx
Sehingga: dx = du /(cocs x)
∫ e^sin x cos x dx = ∫ (e^u cos x)( du/cocs x) = ∫ e^u du = e^u + C = e^sin x + C
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