∫0π/2dx/(1+tanax)=________(其中a为常数).

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问题0π/2dx/(1+tanax)=________(其中a为常数).

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答案π/4

解析 令I=∫0π/2dx/(1+tanax),I=∫0π/2dx/(1+tanax)→∫π/20-dt/(1+cosat)=∫0π/2dx/(1+cosax),则2I=∫0π/2dx/(1+tanax)+∫0π/2dx/(1+cosax)=π/2故∫0π/2dx/(1+tanax)=π/4.
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