∫1/(sinx+cosx)dx上限π/2下限0用万能公式怎么做,
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/07/03 08:41:34
∫1/(sinx+cosx)dx上限π/2下限0用万能公式怎么做,
sinx=2tan(x/2)/[1+tan^2(x/2)]
cosx=[1-tan^2(x/2)]/[1+tan^2(x/2)]
代入得
∫[0,π/2] 1/(sinx+cosx)dx
=∫[0,π/2] [1+tan^2(x/2)]/[1-tan^2(x/2)+2tan(x/2)]dx
=∫[0,π/2] [sec^2(x/2)]/[1-tan^2(x/2)+2tan(x/2)]dx
=∫[0,π/2]1/[1-tan^2(x/2)+2tan(x/2)]dtan(x/2)
=∫[0,π/2]1/[2-(tan^2(x/2)-1)^2]dtan(x/2)
=1/(2√2)∫[0,π/2]{1/[√2-(tan(x/2)-1)]+1/[√2+(tan(x/2)-1)]}dtan(x/2)
=1/(2√2) {-ln[√2-(tan(x/2)-1)]+ln[√2+(tan(x/2)-1)]} [0,π/2]
=1/(2√2) ln[(√2+1)/(√2-1)]
=1/√2ln(√2+1)
再问: 过程倒数第3步一直到最后一步是怎么变的,求详细解释一下
再答: 就是拆项啊
cosx=[1-tan^2(x/2)]/[1+tan^2(x/2)]
代入得
∫[0,π/2] 1/(sinx+cosx)dx
=∫[0,π/2] [1+tan^2(x/2)]/[1-tan^2(x/2)+2tan(x/2)]dx
=∫[0,π/2] [sec^2(x/2)]/[1-tan^2(x/2)+2tan(x/2)]dx
=∫[0,π/2]1/[1-tan^2(x/2)+2tan(x/2)]dtan(x/2)
=∫[0,π/2]1/[2-(tan^2(x/2)-1)^2]dtan(x/2)
=1/(2√2)∫[0,π/2]{1/[√2-(tan(x/2)-1)]+1/[√2+(tan(x/2)-1)]}dtan(x/2)
=1/(2√2) {-ln[√2-(tan(x/2)-1)]+ln[√2+(tan(x/2)-1)]} [0,π/2]
=1/(2√2) ln[(√2+1)/(√2-1)]
=1/√2ln(√2+1)
再问: 过程倒数第3步一直到最后一步是怎么变的,求详细解释一下
再答: 就是拆项啊
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