Question

用分部积分法求以下不定积分:(1)∫1到2 (3x-1)^3+dx (2)∫0到π/2 (sinx)^5+cos (3)∫0到π e^x cosx dx (4)∫0到π x cosx dx 把答案过程写纸上

Answer 1

问：用分部积分法求以下不定积分:(1)∫1到2 (3x-1)^3+dx (2)∫0到π/2 (sinx)^5+cos (3)∫0到π e^x cosx dx (4)∫0到π x cosx dx 把答案过程写纸上

答：(1) 将 (3x-1)^3 视为一整体进行分部积分：设 f(x) = (3x-1)^3，g(x) = x，则有：∫f(x)g'(x)dx = f(x)g(x) - ∫f'(x)g(x)dx∫(3x-1)^3dx = (3x-1)^4/12 + C所以：∫1到2 (3x-1)^3+dx = [(3x-1)^4/12]1到2 = (80/3)。(2) 将 cos 视为一整体，注意 sin(x) 和 cos(x) 的导数轮流出现，需要两次分部积分：设 f(x) = (sinx)^5，g'(x) = cosx，则有：∫f(x)g'(x)dx = f(x)g(x) - ∫f'(x)g(x)dx对于 ∫(sinx)^5cosxdx，令 f(x) = (sinx)^5，g'(x) = cosx，则有：∫f(x)g'(x)dx = f(x)g(x) - ∫f'(x)g(x)dx对于 f'(x)，使用公式(sin^n(x))' = n(sin^(n-1)(x))cos(x)，有：f'(x) = 5(sin^4(x))(cosx)再使用公式 ∫(sin^2(x))^ndx = [(n-1)/n]∫(sin^2(x))^(n-2)dx - [(n-1)/n]∫(sin^2(x))^(n)dx，有：∫(sinx)^5cosxdx = [(sin^6(x))/6] - [(5/6)∫(sin^2(x))^2dx]∫(sin^2(x))^2dx = [(1/2)∫(1-cos(2x))dx]^2 = [(1/2)(x-(1/2)sin(2x))]^2所以：∫0到π/2 (sinx)^5+cosxdx = [(sin^6(x))/6 + (cos(x))]0到π/2 = (2/3)。(3) 令 f(x) = e^x，g'(x) = cosx，则有：∫f(x)g'(x)dx = f(x)g(x) - ∫f'(x)g(x)dx对于 ∫e^xcosxdx，有：∫e^xcosxdx = (1/2)∫e^x(d/dx(sin(x))-cos(x))dx使用分部积分公式，有：∫e^xcosxdx = (1/2)[e^xsinx - ∫e^xsin(x)dx - ∫e^xcos(x)dx]再使用一次分部积分，有：∫e^xcosxdx = (1/2)[e^xs