作业帮证明∫[π/10,x]sint²dt+∫[π/2,x]1/sint²dt=0在（π/10,π/2）内有唯一实根
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![证明∫[π/10,x]sint²dt+∫[π/2,x]1/sint²dt=0在（π/10,π/2）内有唯一实根](/uploads/image/z/15241350-30-0.jpg?t=%E8%AF%81%E6%98%8E%E2%88%AB%5B%CF%80%2F10%2Cx%5Dsint%26%23178%3Bdt%2B%E2%88%AB%5B%CF%80%2F2%2Cx%5D1%2Fsint%26%23178%3Bdt%3D0%E5%9C%A8%EF%BC%88%CF%80%2F10%2C%CF%80%2F2%EF%BC%89%E5%86%85%E6%9C%89%E5%94%AF%E4%B8%80%E5%AE%9E%E6%A0%B9)
证明∫[π/10,x]sint²dt+∫[π/2,x]1/sint²dt=0在（π/10,π/2）内有唯一实根
证明∫[π/10,x]sint²dt+∫[π/2,x]1/sint²dt=0在（π/10,π/2）内有唯一实根
证明∫[π/10,x]sint²dt+∫[π/2,x]1/sint²dt=0在（π/10,π/2）内有唯一实根
令f(x)=∫[π/10,x]sint²dt+∫[π/2,x]1/sint²dt
在（π/10,π/2）
f'(x)=sin^2x+1/sin^2x>0
说明函数在（π/10,π/2）单增
f(π/10)=∫[π/2,π/10]1/sint²dt0
由介值定理得
f(x)在（π/10,π/2）内有唯一实根
即
∫[π/10,x]sint²dt+∫[π/2,x]1/sint²dt=0在（π/10,π/2）内有唯一实根