作业帮高数（分部积分法）问题= =1 ∫e^x乘以(cosx)^2 dx2 ∫e^arcsinx dx3 ∫(sinx)^6 dx 区间-π/2到π/2
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/03 11:39:22
高数（分部积分法）问题= =1 ∫e^x乘以(cosx)^2 dx2 ∫e^arcsinx dx3 ∫(sinx)^6 dx 区间-π/2到π/2
高数（分部积分法）问题= =
1 ∫e^x乘以(cosx)^2 dx
2 ∫e^arcsinx dx
3 ∫(sinx)^6 dx 区间-π/2到π/2
高数（分部积分法）问题= =1 ∫e^x乘以(cosx)^2 dx2 ∫e^arcsinx dx3 ∫(sinx)^6 dx 区间-π/2到π/2
1.cos²x = 1/2 + cos(2x) /2
∫ e^x [1/2 + cos(2x) /2] dx = (1/2) e^x + (1/2) ∫ e^x cos(2x) dx
I = ∫ e^x cos(2x) dx = ∫ cos(2x) d（e^x ) = e^x cos(2x) - ∫ e^x [-2 sin(2x)] dx
= e^x cos(2x) + ∫ 2 sin(2x) d(e^x)
= e^x cos(2x) + 2e^x sin(2x) - 4 ∫ e^x cos(2x) dx
=> I = (1/5) [e^x cos(2x) + 2e^x sin(2x) ] + C
原式 = (1/2) e^x + (1/10) [e^x cos(2x) + 2e^x sin(2x) ] + C
2.I = x * e^arcsinx - ∫ e^arcsinx * x/√(1-x²) dx = x * e^arcsinx + ∫ e^arcsinx d√(1-x²)
= x * e^arcsinx + √(1-x²) e^arcsinx - I
=> I = (1/2) e^arcsinx [ x + √(1-x²) ] + C
3.∫ [-π/2,π/2] (sinx)^6 dx = 2 ∫ [0,π/2] (sinx)^6 dx 有递推公式
= 2 * [ 5*3*1 /(6*4*2)] * π/2
= π/16
1.原式=(1/2)∫(e^x+e^xcos2x)dx
=(1/5)e^xsin2x+(1/2)e^xcos2x+(1/2)∫e^xdx
=(1/5)e^xsin2x+(1/2)e^xcos2x+(1/2)e^x+C
2.u=arcsinx
原式=∫e^ucosudu
=(1/2)(e^usinu+e^ucosu)+C
=(1/2)e^arcsinx...
全部展开
1.原式=(1/2)∫(e^x+e^xcos2x)dx
=(1/5)e^xsin2x+(1/2)e^xcos2x+(1/2)∫e^xdx
=(1/5)e^xsin2x+(1/2)e^xcos2x+(1/2)e^x+C
2.u=arcsinx
原式=∫e^ucosudu
=(1/2)(e^usinu+e^ucosu)+C
=(1/2)e^arcsinx(√(1-x^2)+x)+C
3.可以用公式
∫sin^m(x)dx=-cosxsin^(m-1)(x)/m+(m-1)/m∫sin^(m-2)(x)dx
一步步解决
最后结果是(60x-45sin2x+9sin4x-sin6x)/192+C
原式=π/16
收起
如下图