作业帮求定积分,[-π/2,π/2],((cosx)^(1/2))乘以（sinx）的绝对值 dx我知道（sinx）的绝对值要分[-π/2,0],[0,π/2]两种情况,然后各自还原到原函数F(x),分别是(2/3)(cosx)^(2/3),(-2/3)(cosx)^(2/3）,然后我不知道用哪个F
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![求定积分,[-π/2,π/2],((cosx)^(1/2))乘以（sinx）的绝对值 dx我知道（sinx）的绝对值要分[-π/2,0],[0,π/2]两种情况,然后各自还原到原函数F(x),分别是(2/3)(cosx)^(2/3),(-2/3)(cosx)^(2/3）,然后我不知道用哪个F](/uploads/image/z/1739599-7-9.jpg?t=%E6%B1%82%E5%AE%9A%E7%A7%AF%E5%88%86%2C%5B-%CF%80%2F2%2C%CF%80%2F2%5D%2C%28%28cosx%29%5E%281%2F2%29%29%E4%B9%98%E4%BB%A5%EF%BC%88sinx%EF%BC%89%E7%9A%84%E7%BB%9D%E5%AF%B9%E5%80%BC+dx%E6%88%91%E7%9F%A5%E9%81%93%EF%BC%88sinx%EF%BC%89%E7%9A%84%E7%BB%9D%E5%AF%B9%E5%80%BC%E8%A6%81%E5%88%86%5B-%CF%80%2F2%2C0%5D%2C%5B0%2C%CF%80%2F2%5D%E4%B8%A4%E7%A7%8D%E6%83%85%E5%86%B5%2C%E7%84%B6%E5%90%8E%E5%90%84%E8%87%AA%E8%BF%98%E5%8E%9F%E5%88%B0%E5%8E%9F%E5%87%BD%E6%95%B0F%28x%29%2C%E5%88%86%E5%88%AB%E6%98%AF%282%2F3%29%28cosx%29%5E%282%2F3%29%2C%28-2%2F3%29%28cosx%29%5E%282%2F3%EF%BC%89%2C%E7%84%B6%E5%90%8E%E6%88%91%E4%B8%8D%E7%9F%A5%E9%81%93%E7%94%A8%E5%93%AA%E4%B8%AAF)
求定积分,[-π/2,π/2],((cosx)^(1/2))乘以（sinx）的绝对值 dx我知道（sinx）的绝对值要分[-π/2,0],[0,π/2]两种情况,然后各自还原到原函数F(x),分别是(2/3)(cosx)^(2/3),(-2/3)(cosx)^(2/3）,然后我不知道用哪个F
求定积分,[-π/2,π/2],((cosx)^(1/2))乘以（sinx）的绝对值 dx
我知道（sinx）的绝对值要分[-π/2,0],[0,π/2]两种情况,然后各自还原到原函数F(x),分别是(2/3)(cosx)^(2/3),(-2/3)(cosx)^(2/3）,然后我不知道用哪个F(x)减哪个F(x)?最后答案是4/3
求定积分,[-π/2,π/2],((cosx)^(1/2))乘以（sinx）的绝对值 dx我知道（sinx）的绝对值要分[-π/2,0],[0,π/2]两种情况,然后各自还原到原函数F(x),分别是(2/3)(cosx)^(2/3),(-2/3)(cosx)^(2/3）,然后我不知道用哪个F
考虑曲线y=√(cosx) * sinx
在[-π/2,0],曲线在x轴下,在[0,π/2],曲线在x轴上
∴∫[-π/2,π/2] √(cosx) * |sinx| dx
= -∫[-π/2,0] √(cosx) * sinx dx + ∫[0,π/2] √(cosx) * sinx dx,曲线在x轴下,需加上负号
= ∫[-π/2,0] √(cosx) d(cosx) - ∫[0,π/2] √(cosx) d(cosx)
= (2/3)[cosx]^(3/2)[-π/2,0] - (2/3)[cosx]^(3/2)[0,π/2]
= (2/3)(1-0) - (2/3)(0-1)
= 2/3 + 2/3
= 4/3