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求∫[0π/4](1-sin2x)/(1+sin2x)dx
题目详情
求∫[0 π/4](1-sin2x)/(1+sin 2x)dx
▼优质解答
答案和解析
(1-sin2x)/(1+sin 2x)=（sinx-cosx）^2/(sinx+cosx)^2=[(tanx-1)/(tanx+1)]^2=tan(x-π/4)^2
=sec(x-π/4)^2-1;∫[0 π/4](1-sin2x)/(1+sin 2x)dx=tan(x-π/4)-x+C
=sec(x-π/4)^2-1;∫[0 π/4](1-sin2x)/(1+sin 2x)dx=tan(x-π/4)-x+C
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