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
for a more clear author please click the link below
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
for a more clear author please click the link below
Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Tentukanlah Integral dari :
-
∫ ctg x dx
-
∫ (x dx) /(1- x^2 )
Jawab:
1. ctg x = cosx/sinx
misalkan: u = sin x
maka: du/dx= cosx
Sehingga: dx = du/cosx
∫ctg x dx = ∫ cosx / sinx dx = ∫( cosx /u) ( du/cosx) = ∫ du / u = ln|u| + C = ln| sinx | + C
2. misalkan: u = 1- x^2
maka: du/dx = -2x
Sehingga: dx = du /(-2x)
∫ (x dx) /(1- x^2 ) = ∫ (x/u) [du /(-2x)] = 1 /(-2) ∫ du /u = 1/(-2) ln |u| + C = 1/(-2) ln |1- x^2 | + C
for a more clear author please click the link below
Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Tentukanlah Integral berikut:
-
∫ cos(2x-1)dx
-
∫ dx/(2x+3)
Jawab:
1. Misalkan: u = 2x-1
Maka: du/dx = 2
Sehingga: dx = du/2
∫ cos(2x-1) dx = ∫ cosu du /2 = 1/2 ∫ cos u du = 1/2 ( sin u ) + C = 1/2 sin(2x-1) + C
2. Misalkan: u = 2x+3
Maka: du/dx = 2
Sehingga: dx = du/2
∫ dx /(2x+3) = ∫ (1 /u) (du/2) = 1/2 ∫ du / u = 1/2 ln|u| + C = 1/2 ln|2x + 3| + C
for a more clear author please click the link below
Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Tentukanlah integral berikut:
-
∫ sin^2 3x cos 3x dx
-
∫ cos x dx / √(sinx )
Jawab:
1. Misalkan: u = sin 3x
Maka: du/dx=3cos3x
Sehingga : dx = du/(3 cos3x )
Atau: cos 3x dx = du/3
∫ sin^2 3x cos 3x dx = ∫ u^2 cos 3x du / (3 cos3x ) = 1/3 ∫ u^2 du = 1/3 1/3 u^3 + c = 1/9 (sin 3x)^3 + C
2. Misalkan: u = sin x
Maka: du/dx = cosx
Sehingga : dx = du/cosx
∫ cos x dx /√(sinx ) = ∫ (cosx ) /(u)^(1/2) du/cosx = ∫ (u)^(-1/(2 )) du = 1/(- 1/2+ 1) u^(1/2) + C = 2 √u + C = 2 √(sinx ) + C
for a more clear author please click the link below