搜搜考试网计算这个积分
来源：学生作业帮 编辑：搜搜考试网作业帮 分类：综合作业 时间：2024/05/07 00:40:09
计算这个积分
∫[-π/4,π/4]x/(1+sinx)dx=∫[-π/4,π/4]x(1-sinx)/cos^2xdx
=∫[-π/4,π/4]xd(tanx-1/cosx)
=x(tanx-1/cosx)|[-π/4,π/4]-∫[-π/4,π/4](tanx-1/cosx)dx
=-π/√2+∫[-π/4,π/4](1-sinx)/cosxdx
=-π/√2+∫[-π/4,π/4]cosx/(1+sinx)dx
=-π/√2+ln(1+sinx)|[-π/4,π/4]
=-π/√2+2ln(√2+1)
=∫[-π/4,π/4]xd(tanx-1/cosx)
=x(tanx-1/cosx)|[-π/4,π/4]-∫[-π/4,π/4](tanx-1/cosx)dx
=-π/√2+∫[-π/4,π/4](1-sinx)/cosxdx
=-π/√2+∫[-π/4,π/4]cosx/(1+sinx)dx
=-π/√2+ln(1+sinx)|[-π/4,π/4]
=-π/√2+2ln(√2+1)