Calculo 1. Alguém por favor poderia me ajudar com pelo menos 1 dessas questoes?????
Soluções para a tarefa
a)
∫cos 2x dx
Fazendo u=2x ==> du =2 dx ==> dx= du/2
∫cos u du/2 = (1/2) * sen u
Como u= 2x ==> (1/2) * sen (2x)
π/2
(1/2) * [ sen 2x] = (1/2) * [ sen π - sen (-π/3)] =(1/2) * √3/2 = √3/4
-π/3
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b)
∫ (1+3x²)/x dx = ∫ 1/x +3x dx = ln x + 3x²/2
2
==> [ln x + 3x²/2] = ln 2 + 6 - ln 1 -3/2 = ln 2 +9/2
1
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c)
∫ (sen x + sen 2x) dx =∫ sen x dx + ∫ sen 2x dx
∫ sen x dx= - cos x
∫ sen 2x dx ...faça u=2x ==> du = dx ==> ∫ sen u du/2
= (1/2) * (-cos u) como u=2x ==> = - (1/2) * cos (2x)
π/3
[ - cos x - (1/2) * cos (2x) ]
0
= - cos π/3 - (1/2) * cos (2π/3) + cos 0 + (1/2)* cos 0
=-1/2 - (1/2) * (-1/2) + 1 + 1/2 =5/4
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d)
∫ sec² x dx
sec² x = 1/cos²x = 1+ sen²x/cos²x = 1 + sen x * sen x/cos²x
∫ 1 + sen x * sen x/cos²x dx
∫ 1 dx + ∫ sen x * sen x/cos²x dx
∫ 1 dx =x
∫ sen x * sen x/cos²x dx
Integrando por partes:
u= sen x ==> du =cos x dx
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dv =sen x/cos²x dx ==> ∫ dv =∫ sen x/cos²x dx
u= cos x ==> du =-sen x dx
∫ sen x/u² du/(-sen x) = - ∫ 1/u² du = u⁻¹/(-1) = 1/u
sendo u = cos x ==> ∫ sen x/cos²x dx = 1/cos x
v = 1/cos x
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∫ sen x * sen x/cos²x dx =(1/cos x) * sen x - ∫ 1/cos x * cos x dx
=tan x -x
∫ sec² x dx = x + tan x -x = tan x
π/4
= [ tan x ] = tan π/4 - tan 0 = 1 - 0 = 1
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