Question

高数求解，急

求详细过程

Answer 1

先求不定积分
设t=2x dt=2dx dx=0.5dt
则上式可表为0.5∫e^tsintdt.........(1)
设u=e^t du=e^tdt dt=du/e^t
dv=sintdt v=cost
∫e^tsintdt=e^tcost-∫coste^tdt=e^tcost+e^tsint-∫e^tsintdt
2∫e^tsint=e^t(sint+cost) 代入(1)式得
∫e^tsintdt=e^t(sint+cost)=e^2x(sin2x+cos2x)代入积分区间则得.

Answer 2

～ ∫ ≠ ≤ ≥ ≈ ∞ ∶ 3运算符号 × ÷ √
∫(0->π/2)e^(2x) .sin(2x) dx
=(1/2)∫(0->π/2)sin(2x) de^(2x)
=(1/2)[sin(2x).e^(2x)](0->π/2) -(1/4)∫(0->π/2)cos(2x) e^(2x)dx
=(1/2)e^π - (1/8)∫(0->π/2)cos(2x) de^(2x)
=(1/2)e^π - (1/8)[cos(2x).e^(2x)](0->π/2) - (1/16)∫(0->π/2)sin(2x).e^(2x)dx
=(1/2)e^π + (1/8)[1+e^π] - (1/16)∫(0->π/2)sin(2x).e^(2x)dx
(15/16)∫(0->π/2)e^(2x) .sin(2x) dx =(5/8)e^π + 1/8
∫(0->π/2)e^(2x) .sin(2x) dx =(16/15) [ (5/8)e^π + 1/8 ]
= (2/3)e^π + 2/15