please show how to calculate definite integral of y=|sin x-square root 3*cos x|, in interval [0,pi/2]?

The absolute value forces the answers to definite
integrals to be positive on the interval [0,pi/2].  Therefore the integral needs to be
split up.  Looking at the graph of sin x - sqrt 3 (cos x), from [0,pi/3] the y-values
are negative. From [pi/3, pi/2] the y-values are
positive.

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sscr="-1,3,-3,3,1,1,1,1,1,300,200,func,sin(x)-sqrt(3)*cos(x),null,0,0,,,black,1,none"/>

The
definite integral breaks into two.

class="AM"> from [0,pi/3]
plus

sin(x)-sqrt(3)cos(x)dx from [pi/3, pi/2]

The
integral is equal to -cos(x)-sqrt(3)*sin(x)

Putting in the
bounds:

-[-cos(pi/3)-sqrt(3)sin(pi/3)-(-cos(0)-sqrt(3)sin(0))]+

[-cos(pi/2)-sqrt(3)sin(pi/2)-(-cos(pi/3)-sqrt(3)sin(pi/3))]
=

-[-1/2-3/2-(-1-0)] + [-0-sqrt(3)-(-1/2-3/2)]
=

-[-1]+[2-sqrt(3)] =
3-sqrt(3)

The exact answer is 3-sqrt(3) which is
approximately 1.268.