Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Tentukanlah Integral berikut:
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∫ cos(2x-1)dx
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∫ 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
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Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Tentukanlah integral berikut:
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∫ sin^2 3x cos 3x dx
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∫ 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
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Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Hitunglah integral berikut:
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∫ (2x+10)dx
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∫ √(8+5x) dx
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∫ dx /(3x+1)^2
Jawab:
1.∫ (2x+10) dx = 2/(1+1) x^(1+1)+ 10x+C = x^2+ 10x+C
2. Misalkan : u = 8 + 5x
Maka: du/dx=5 Sehingga: dx = 1/5 du
∫ √(8+5x) dx = ∫ (u)^(1/2) 1/5 du = 1/5 ∫ (u)^(1/2) du = (1/5) 1/(1/2+ 1) u^(1/2+1)+ C = (1/5) 2/3 u^(3/2)+ C = 2/15 (8+5x)^(3/2)+ C= 2/15 (8+5x) √(8+5x)+C
3. Misalkan: u = 3x+1
Maka: du/dx=3 Sehingga : dx = du/3
∫ dx /(3x+1)^2 = ∫ (du/3) /(u)^2 = 1/3 ∫ du /(u)^2 = 1/3 1/(-2+1) (u)^(-2+1) +C = 1 /(-3) u^(-1) + C = 1 / (-3) (3x+1)^(-1) + C = (-1) /(9x+3) + C
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Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Rumus lengkap Integral tak tentu
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∫du = u+c
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∫a du = a ∫du
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∫(du+dv) = ∫du+ ∫dv
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∫(du-dv) = ∫du- ∫dv
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∫u^n du = u^n/(n+1)+ C ,n tidak sama dengan-1
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∫du/u = ln| u | + C
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∫e^u du = e^u+ c
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∫a^u du = a^u / lna + C ,a>0 dan a ≠1
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∫√(a^2- x^2 ) dx = x/2 √(a^2- x^2 )+ a^2/2 arc sin (x/a)+ C
Rumus Integral untuk fungsi Trigonometri
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∫sin u du = – cos u+C
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∫cos u du = sin u+C
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∫ sec^2 u du = tg u+C
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∫ cosec^2 u du = – ctg u+C
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∫ sec u tg u du = sec u+C
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∫ cosec u ctg u du = – cosec u+C
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∫ sin^n ax cos ax dx = 1/a ∫ u^n du = 1/a u^(n+1)/(n+1)+C ,n ≠ -1 dan u=sinax
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∫ sin^n ax cos ax dx = 1/a ∫ u^n du = 1/a ln| u |+C ,n= -1 dan u = sinax
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∫ sin ^n u du = – (sin^(n-1) u cosu)/n+ (n-1)/n ∫ sin^(n-2) u du+c
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∫ cos ^n u du= (cos^(n-1) u sin u)/n+ (n-1)/n ∫ cos^(n-2) u du+c
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∫ sin ax sin bx dx = 1/2 [ sin(a-b)x/(a-b)- (sin ( a+b)x)/(a+b)]+C
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∫ sin ax cos bx dx = -1/2 [ cos(a-b)x/(a-b)+ (cos ( a+b)x)/(a+b)]+C
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∫ cos ax cos bx dx = 1/2 [ sin(a-b)x/(a-b)+ (sin ( a+b)x)/(a+b)]+C
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∫ sin^2 ax dx = 1/2 (x-sin2ax/2a) +C
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∫ cos^2 ax dx = 1/2 (x+ sin2ax/2a) +C
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∫ x^n sin ax dx = (-1) / a x^n cos ax+ n/a ∫ x^(n-1) cos ax dx
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∫ x^n cos ax dx = 1 / a x^n sin ax+ n/a ∫ x^(n-1) sin ax dx
Rumus Integral dari Fungsi invers trigonometri
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∫ du/√(1- u^2 ) = arc sin u+C
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∫ -du/√(1- u^2 ) = arc cos u+C
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∫ du/(1+u^2 ) = arc tg u+C
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∫ -du/(1+u^2 ) = arc ctg u+C
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∫ du/(u √( u^2- 1)) = arc sec |u|+C
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∫ -du/(u √( u^2- 1)) = arc cosec |u|+C
Rumus integral Parsial
∫ v du= uv- ∫ u dv
Rumus Integral Rangkap dua melalui daerah tertutup s
1. ∬ f(x,y) dxdy = ∫_(b_1)^(b_2) ∫_(x_1)^(x_2) f(x,y) dxdy
2. ∬f(x,y) dxdy = ∫_(a_1)^(a_2) ∫_(y_1)^(y_2) f(x,y) dxdy
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Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Tentukanlah Turunan pertama dari fungsi hiperbol trigonometri berikut!
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y = x sinh x
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y = x^2/cosh x
Jawab:
1. dy/dx=1(sinh x) – x d/dx(sinh x)
dy/dx = sinh x + x cosh x
Atau
dy/dx = (e^x- e^(-x) )/2+ x ((e^x+ e^(-x) )/2)
2. dy/dx = (2x(cosh x) – x^2 (sinh x)) / (cosh x)^2
Atau :
dy/dx = (2x((e^x+ e^(-x) )/2)-x^2 ((e^x- e^(-x) )/2) )/((e^x+ e^(-x) )/2)^2
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