定积分在区间[0,π/2]∫[1/1+(tanx)^√2]dx
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解决时间 2021-12-15 13:39
- 提问者网友：喜遇你
- 2021-12-14 23:30
定积分在区间[0,π/2]∫[1/1+(tanx)^√2]dx
最佳答案
- 二级知识专家网友：哥在撩妹请勿打扰
- 2021-12-15 00:08
注意一个结论：∫[0,π/2] f(sinx)dx=∫[0,π/2] f(cosx)dx （定积分换元法那里的一道例题）
则 ∫[0,π/2] f(sinx)dx=1/2[∫[0,π/2] f(sinx)dx+∫[0,π/2] f(cosx)dx]
∫[0,π/2][1/1+(tanx)^√2]dx
=∫[0,π/2][(cosx)^√2/[(cosx)^√2+(sinx)^√2]dx
=1/2{ ∫ [0,π/2][(cosx)^√2/[(cosx)^√2+(sinx)^√2]dx+∫[0,π/2] [(sinx)^√2/[(cosx)^√2+(sinx)^√2]dx }
=1/2∫[0,π/2] 1dx
=π/4
则 ∫[0,π/2] f(sinx)dx=1/2[∫[0,π/2] f(sinx)dx+∫[0,π/2] f(cosx)dx]
∫[0,π/2][1/1+(tanx)^√2]dx
=∫[0,π/2][(cosx)^√2/[(cosx)^√2+(sinx)^√2]dx
=1/2{ ∫ [0,π/2][(cosx)^√2/[(cosx)^√2+(sinx)^√2]dx+∫[0,π/2] [(sinx)^√2/[(cosx)^√2+(sinx)^√2]dx }
=1/2∫[0,π/2] 1dx
=π/4
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