What is #F(x) = int cos^2x-tan^3x+sinx dx# if #F(pi/3) = 1 #?

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Answer 1 · 2016-11-03T08:13:57

#F(x)=1/2x+1/4 sin 2x-1/2 tan^2x-ln cos x-cos x+9/2-1/8sqrt 3- ln 2-pi/6#

Use du=#(du)/(dx)#dx. Now,

#F(x)=int((1+cos 2x)/2-tan x (sec^2x-1)+sin x) dx#

#=intd(1/2x)+int d((1/2((sin 2x)/2))-int d((tan^2x)/2)#

#+int d(-ln (cos x))+int d(-cos x)#

#=1/2x+1/4 sin 2x-1/2 tan^2x-ln cos x-cos x + C#

#F(pi/3)=1/6pi+1/4 sin (2/3pi)-tan^2(1/3pi)-ln (cos (1/3pi)-cos (1/3pi) +C#

#=1/6pi+1/8sqrt 3 - 3-ln(1/2)-1/2+C#

#=1#

So, #C=9/2-1/8sqrt 3- ln 2-pi/6#