Question

∫e^xsinxdx,(0-π/2)；∫(x^5+x^3-x+1)sinx^2dx,(-π/4，π/4），求定积分，求过程，谢谢。

Answer 1

∫(0→π/2) e^xsinx dx = - ∫(0→π/2) e^x dcosx
= - e^xcosx：(0→π/2) + ∫(0→π/2) e^xcosx dx，分部积分法
= 1 + ∫(0→π/2) e^x dsinx
= 1 + e^xsinx：(0→π/2) - ∫(0→π/2) e^xsinx dx
2∫(0→π/2) e^xsinx dx = 1 + e^(π/2)
==> ∫(0→π/2) e^xsinx dx = [1 + e^(π/2)]/2

∫(-π/4→π/4) (x⁵ + x³ - x + 1)sin²x dx
= ∫(-π/4→π/4) x⁵sin²x dx + ∫(-π/4→π/4) x³sin²x - ∫(-π/4→π/4) xsin²x + ∫(-π/4→π/4) sin²x dx
= 0 + 0 - 0 + 2∫(0→π/4) sin²x dx，前面三项奇函数，最后一项偶函数
= ∫(0→π/4) (1 - cos2x) dx
= x - (1/2)sin2x：(0→π/4)
= π/4 - 1/2