证明∫sinx/sinx+cosxdx=∫cosx/sinx+cosxdx=π/4 ,积分上限是π/2,下限是0
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证明∫sinx/sinx+cosxdx=∫cosx/sinx+cosxdx=π/4 ,积分上限是π/2,下限是0
证明∫sinx/sinx+cosxdx=∫cosx/sinx+cosxdx=π/4 ,积分上限是π/2,下限是0
证明∫sinx/sinx+cosxdx=∫cosx/sinx+cosxdx=π/4 ,积分上限是π/2,下限是0
换元法令t=(π/2) -x,那么x=(π/2)-t,dx=-dt,积分限t上限是0,下限是(π/2)
∫[0,π/2] [sinx/(sinx+cosx)]dx=
∫[π/2,0] sin((π/2)-t)/(sin((π/2)-t)+cos((π/2)-t))d((π/2)-t)
=∫[π/2,0] -sint/(cost+sint)dt
=∫[0,π/2] c...
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换元法令t=(π/2) -x,那么x=(π/2)-t,dx=-dt,积分限t上限是0,下限是(π/2)
∫[0,π/2] [sinx/(sinx+cosx)]dx=
∫[π/2,0] sin((π/2)-t)/(sin((π/2)-t)+cos((π/2)-t))d((π/2)-t)
=∫[π/2,0] -sint/(cost+sint)dt
=∫[0,π/2] cost/sint+cost dt
=∫[0,π/2] cosx/(sinx+cosx) dx
又因为∫[0,π/2] [sinx/(sinx+cosx)]dx+∫[0,π/2] cosx/(sinx+cosx) dx
=∫[0,π/2] 1 ×dx
=π/2
所以
∫[0,π/2] sinx/(sinx+cosx)dx=∫cosx/(sinx+cosx)dx=π/4
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