∫1⼀(sinx+cosx)dx上限π⼀2下限0用万能公式怎么做，要详细过程谢谢。

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)

积化和差公式：

sinα·cosβ=(1/2)[sin(α+β)+sin(α-β)]

cosα·sinβ=(1/2)[sin(α+β)-sin(α-β)]

cosα·cosβ=(1/2)[cos(α+β)+cos(α-β)]

sinα·sinβ=-(1/2)[cos(α+β)-cos(α-β)]

和差化积公式：

sinα+sinβ=2sin[(α+β)/2]cos[(α-β)/2]

sinα-sinβ=2cos[(α+β)/2]sin[(α-β)/2]

cosα+cosβ=2cos[(α+β)/2]cos[(α-β)/2]

cosα-cosβ=-2sin[(α+β)/2]sin[(α-β)/2]

同角三角函数的基本关系式

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)