搜搜考试网f(x)=∫(0-x)cost/1+sint²则∫（0-π/2）f'(x)/1+f²(x) dx =
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f(x)=∫(0-x)cost/1+sint²则∫（0-π/2）f'(x)/1+f²(x) dx =
希望详细易懂 本人数学不好
希望详细易懂 本人数学不好
f(x)=∫(0,x) cost / 1+sin^2t dt
则
∫(0,π/2) f'(x) / 1+f^2(x) dx
=∫(0,π/2) 1 / 1+f^2(x) d(f(x))
=arctan(f(x)) | (0,π/2)
=arctan(f(π/2)) - arctan(f(0))
又有：
f(0)=∫(0,0) cost / 1+sin^2t dt=0,积分上下限一样,积分明显为0
f(π/2)
=∫(0,π/2) cost / 1+sin^2t dt
=∫(0,π/2) 1 / 1+sin^2t d(sint)
=arctan(sint) | (0,π/2)
=arctan(1) - arctan(0)
=π/4-0
=π/4
因此,
∫(0,π/2) f'(x) / 1+f^2(x) dx
=arctan(f(π/2)) - arctan(f(0))
=π/4-0
=π/4
有不懂欢迎追问
则
∫(0,π/2) f'(x) / 1+f^2(x) dx
=∫(0,π/2) 1 / 1+f^2(x) d(f(x))
=arctan(f(x)) | (0,π/2)
=arctan(f(π/2)) - arctan(f(0))
又有：
f(0)=∫(0,0) cost / 1+sin^2t dt=0,积分上下限一样,积分明显为0
f(π/2)
=∫(0,π/2) cost / 1+sin^2t dt
=∫(0,π/2) 1 / 1+sin^2t d(sint)
=arctan(sint) | (0,π/2)
=arctan(1) - arctan(0)
=π/4-0
=π/4
因此,
∫(0,π/2) f'(x) / 1+f^2(x) dx
=arctan(f(π/2)) - arctan(f(0))
=π/4-0
=π/4
有不懂欢迎追问
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