求定积分∫[-π/2~π/2][sinx/1+x^2+(cosx)^2]dx
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求定积分∫[-π/2~π/2][sinx/1+x^2+(cosx)^2]dx
![求定积分∫[-π/2~π/2][sinx/1+x^2+(cosx)^2]dx](/uploads/image/z/7721294-14-4.jpg?t=%E6%B1%82%E5%AE%9A%E7%A7%AF%E5%88%86%E2%88%AB%5B-%CF%80%2F2%7E%CF%80%2F2%5D%5Bsinx%2F1%2Bx%5E2%2B%28cosx%29%5E2%5Ddx)
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
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
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