moise-gunen
moise-gunen
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f(x)=(senx +cosx)^2 = sen^2 x + cos^2 x +2sen x cos x = 1+sin(2x) O periodo da funçao T 1+sen(2x)= 1+sen(2(x+T)) 2x+2pi=2x+2T T = pi conjunto imagem -1<=sen(2x)<=1 Adicione 1 a todos os membros das desigualdades 0<=1+sen(2x)&…
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f(x)=(senx +cosx)^2 = sen^2 x + cos^2 x +2sen x cos x = 1+sin(2x) O periodo da funçao T 1+sen(2x)= 1+sen(2(x+T)) 2x+2pi=2x+2T T = pi conjunto imagem -1<=sen(2x)<=1 Adicione 1 a todos os membros das desigualdades 0<=1+sen(2x)&…
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∫1/[(2+cosx)*sinx] dx = ∫sin(x) /[(2+cos(x))sin^2(x)] dx = - ∫ 1/[(2+cos(x))(1-cos^2(x)] d(cos(x)) cos(x) = u - ∫ 1/[(2 + u)(1-u^2)] du = A∫1/(2+u)du +B∫1/(1+u) du + C∫1/(1-u) du A(1-u^2) + B(1-u)(2+u) + C(2+u)(1+u) = - 1 C = -1/6 B…
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∫1/[(2+cosx)*sinx] dx = ∫sin(x) /[(2+cos(x))sin^2(x)] dx = - ∫ 1/[(2+cos(x))(1-cos^2(x)] d(cos(x)) cos(x) = u - ∫ 1/[(2 + u)(1-u^2)] du = A∫1/(2+u)du +B∫1/(1+u) du + C∫1/(1-u) du A(1-u^2) + B(1-u)(2+u) + C(2+u)(1+u) = - 1 C = -1/6 B…
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|x² -2x +3| = x² -2x +3 Because x² -2x +3 > 0 for any real x. (delta<0 and coef of x² it's positive) Then you must solve x² -2x +3≤ 4 move 4 with changed sign from right to left x² -2x -1≤ 0 x² -2x +1-2≤ 0 (x-1)² ≤ 2 then -sqrt(…
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you must power 5 ^5 =3125 Answer 3125
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you must power 5 ^5 =3125 Answer 3125
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you must power 5 ^5 =3125 Answer 3125
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∫1/[(2+cosx)*sinx] dx = ∫sin(x) /[(2+cos(x))sin^2(x)] dx = - ∫ 1/[(2+cos(x))(1-cos^2(x)] d(cos(x)) cos(x) = u - ∫ 1/[(2 + u)(1-u^2)] du = A∫1/(2+u)du +B∫1/(1+u) du + C∫1/(1-u) du A(1-u^2) + B(1-u)(2+u) + C(2+u)(1+u) = - 1 C = -1/6 B…