What is #f(x) = int sinx-cos^2x dx# if #f(pi/6)=2 #?

1 Answer
maganbhai P.
Mar 10, 2018

#f(x)=-[cosx+x/2+1/4*sin2x]+(48+15sqrt(3)+2pi)/24#

Explanation:

#f(x)=int(sinx-cos^2x)dx#
#f(x)=int(sinx-1/2(1+cos2x))dx=[-cosx-1/2(x+(sin2x)/2)]+c#
#f(x)=[-cosx-x/2-1/4*sin2x]+c#, #to (1)#
#f(pi/6)=[-cos(pi/6)-pi/12-1/4*sin(pi/3)]+c=2=>[-sqrt(3)/2-pi/12-1/4sqrt(3)/2]+c=2=>-[sqrt(3)/2+pi/12+1/4sqrt(3)/2]+c=2##=>-(12sqrt(3)+2pi+3sqrt(3))/24+c=2=>c=2+(15sqrt(3)+2pi)/24#
#.c=(48+15sqrt(3)+2pi)/24#, #to (2)#
From (1) and (2)
#f(x)=-[cosx+x/2+1/4*sin2x]+(48+15sqrt(3)+2pi)/24#

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