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
Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Hitunglah integral berikut:
-
∫ (2x+10)dx
-
∫ √(8+5x) dx
-
∫ 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
for a more clear author please click the link below
Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Rumus lengkap Integral tak tentu
-
∫du = u+c
-
∫a du = a ∫du
-
∫(du+dv) = ∫du+ ∫dv
-
∫(du-dv) = ∫du- ∫dv
-
∫u^n du = u^n/(n+1)+ C ,n tidak sama dengan-1
-
∫du/u = ln| u | + C
-
∫e^u du = e^u+ c
-
∫a^u du = a^u / lna + C ,a>0 dan a ≠1
-
∫√(a^2- x^2 ) dx = x/2 √(a^2- x^2 )+ a^2/2 arc sin (x/a)+ C
Rumus Integral untuk fungsi Trigonometri
-
∫sin u du = – cos u+C
-
∫cos u du = sin u+C
-
∫ sec^2 u du = tg u+C
-
∫ cosec^2 u du = – ctg u+C
-
∫ sec u tg u du = sec u+C
-
∫ cosec u ctg u du = – cosec u+C
-
∫ sin^n ax cos ax dx = 1/a ∫ u^n du = 1/a u^(n+1)/(n+1)+C ,n ≠ -1 dan u=sinax
-
∫ sin^n ax cos ax dx = 1/a ∫ u^n du = 1/a ln| u |+C ,n= -1 dan u = sinax
-
∫ sin ^n u du = – (sin^(n-1) u cosu)/n+ (n-1)/n ∫ sin^(n-2) u du+c
-
∫ cos ^n u du= (cos^(n-1) u sin u)/n+ (n-1)/n ∫ cos^(n-2) u du+c
-
∫ sin ax sin bx dx = 1/2 [ sin(a-b)x/(a-b)- (sin ( a+b)x)/(a+b)]+C
-
∫ sin ax cos bx dx = -1/2 [ cos(a-b)x/(a-b)+ (cos ( a+b)x)/(a+b)]+C
-
∫ cos ax cos bx dx = 1/2 [ sin(a-b)x/(a-b)+ (sin ( a+b)x)/(a+b)]+C
-
∫ sin^2 ax dx = 1/2 (x-sin2ax/2a) +C
-
∫ cos^2 ax dx = 1/2 (x+ sin2ax/2a) +C
-
∫ x^n sin ax dx = (-1) / a x^n cos ax+ n/a ∫ x^(n-1) cos ax dx
-
∫ 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
-
∫ du/√(1- u^2 ) = arc sin u+C
-
∫ -du/√(1- u^2 ) = arc cos u+C
-
∫ du/(1+u^2 ) = arc tg u+C
-
∫ -du/(1+u^2 ) = arc ctg u+C
-
∫ du/(u √( u^2- 1)) = arc sec |u|+C
-
∫ -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
for a more clear author please click the link below