Rumus lengkap Integral tak tentu
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