∫(cosx)^3 *xdx

Question

如题所述

Answer 1

用局部代换 u'v = uv - ∫uv'
∫(cosx)^3 *xdx
= ∫(cos(x))^2 *x d(sin(x))
=∫(1 -sin^2(x)) *x d(sin(x))
= (sin(x) - 1/3 * sin^3(x)) * x - ∫(sin(x) - 1/3 * sin^3(x)) dx
= x * sin(x) - 1/3 * x * sin^3(x) + cos (x) + ∫(1/3) * sin^3(x) dx
= x * sin(x) - 1/3 * x * sin^3(x) + cos (x) - (1/3) ∫sin^2(x) d(cos(x))
= x * sin(x) - 1/3 * x * sin^3(x) + cos (x) - 1/3 * cos(x) + 1/ 9 * (cos^2(x)) + C
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Answer 2

结果为x*(1/3*cos(x)^2*sin(x)+2/3*sin(x))+1/9*cos(x)^3+2/3*cos(x)+C

Answer 3

∫vdu'= uv - ∫udv'
∫(cosx)^3 *xdx
= ∫(cos(x))^2 *x d(sin(x))
=∫(1 -sin^2(x)) *x d(sin(x))
= ∫xd(sin(x) - ∫ x*sin^2(x)d(sin(x))
= x * sin(x) - ∫sin(x)dx -∫ x*d(sin^3(x) /3)
= x * sin(x) + cos (x) - x*sin^3(x) /3+∫ sin^3(x) /3dx
= x * sin(x)+ cos (x) -x*sin^3(x) /3-∫ (1-cos ^2(x))/3dcos (x)
=x * sin(x)+ cos (x) -x*sin^3(x) /3- cos (x) /3+cos ^3(x)/9+C本回答被提问者采纳