求定积分∫(0到π/2)[cos(x/2)-sin(x/2)]^2dx运用到什么三角变换?

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求定积分∫(0到π/2)[cos(x/2)-sin(x/2)]^2dx运用到什么三角变换?
求定积分∫(0到π/2)[cos(x/2)-sin(x/2)]^2dx
运用到什么三角变换?

求定积分∫(0到π/2)[cos(x/2)-sin(x/2)]^2dx运用到什么三角变换?
[cos(x/2)-sin(x/2)]²
=[cos²(X/2)+sin²(x/2)]+2sin(x/2)cos(x/2)
=1+sinx
∫(π/2,0)[cos(x/2)-sin(x/2)]^2dx
=∫(π/2,0) 1+sinx dx
=x|(π/2,0)-cosx|(π/2,0)
=(π/2)+1

=0到二分之派       1-2sinx/2*cosx/2   dx=~~~~~~~~~~~       1-sinx  dx=x+cosx     /0到二分之派=0+1-0-二分之=1-二分之派