Question

求定积分∫[-π/2~π/2][sinx/1+x^2+(cosx)^2]dx

Answer 1

如果是∫(-π/2~π/2) sinx/(1 + x² + cos²x) dx，分子奇函数，分母偶函数，整式是奇函数
所以该定积分等于0
如果是：
∫(-π/2~π/2) [sinx/(1 + x²) + cos²x] dx
= ∫(-π/2~π/2) sinx/(1 + x²) dx + ∫(-π/2~π/2) cos²x dx，前面一项奇函数，后面一项偶函数
= 0 + 2∫(0~π/2) cos²x dx
= 2∫(0~π/2) (1 + cos2x)/2 dx
= x + 1/2 * sin2x |(0~π/2)
= π/2