∫1/(sinx+cosx)dx上限π/2下限0用万能公式怎么做,
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∫1/(sinx+cosx)dx上限π/2下限0用万能公式怎么做,
![∫1/(sinx+cosx)dx上限π/2下限0用万能公式怎么做,](/uploads/image/z/15867607-31-7.jpg?t=%E2%88%AB1%2F%28sinx%2Bcosx%29dx%E4%B8%8A%E9%99%90%CF%80%2F2%E4%B8%8B%E9%99%900%E7%94%A8%E4%B8%87%E8%83%BD%E5%85%AC%E5%BC%8F%E6%80%8E%E4%B9%88%E5%81%9A%2C)
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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