Question
Wed November 23, 2011 By: Ankur Jain

Integration of(sin2x-cos2x)dx = (2)^-1/2 sin(2x-a)+b then find a & b

Expert Reply
Wed November 23, 2011
?(sin2x-cos2x)dx = (1/?2)sin(2x-a) + b
 
?sin2xdx-?cos2xdx = (1/?2)sin(2x-a) + b
 
-cos2x/2-sin2x/2+C= sin2xcosa/?2 - cos2xsina/?2 +b
 
Compare coefficients of sin2x cos2x and constant , we get
 
-1/2=cosa/?2 so, cosa= -1/?2
 
-1/2=-sina/?2  So , sina=1/?2
 
Sin is positive and cos is negative means a is in II quadrant
 
hence a=90+sin-11/?2 = 90+45 = 135° = 3?/4 rad
 
and b= C . any constant
 
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