What is #f(x) = int x-sin2x-6cosx dx# if #f(pi/2)=3 #?

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Answer 1 · 2016-12-22T12:40:29

#f(x)=x^2/2+1/2cos2x-6sinx+7/2-pi^2/8#

#f(x)=int(x-sin2x-6cosx)dx##" "f(pi/2)=3#

integrate term by term

#f(x)=x^2/2+1/2cos2x-6sinx+C#

#f(pi/2)=1/2xx(pi/2)^2+1/2cos(2xxpi/2)-6sin(2xxpi/2)+C=3#

#pi^2/8+1/2cancel(cospi)^-1- 6cancel(sinpi)^0+C=3#

#pi^2/8-1/2+C=3#

#C=3+1/2-pi^2/8=7/2-pi^2/8#

#f(x)=x^2/2+1/2cos2x-6sinx+7/2-pi^2/8#