What is F(x) = int cos^2x-tan^3x+sinx dxF(x)=∫cos2x−tan3x+sinxdx if F(pi/3) = 1 F(π3)=1?

1 Answer
A. S. Adikesavan
Nov 3, 2016

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/6F(x)=12x+14sin2x−12tan2x−lncosx−cosx+92−18√3−ln2−π6

Explanation:

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

F(x)=int((1+cos 2x)/2-tan x (sec^2x-1)+sin x) dxF(x)=∫(1+cos2x2−tanx(sec2x−1)+sinx)dx

=intd(1/2x)+int d((1/2((sin 2x)/2))-int d((tan^2x)/2)=∫d(12x)+∫d((12(sin2x2))−∫d(tan2x2)

+int d(-ln (cos x))+int d(-cos x)+∫d(−ln(cosx))+∫d(−cosx)

=1/2x+1/4 sin 2x-1/2 tan^2x-ln cos x-cos x + C=12x+14sin2x−12tan2x−lncosx−cosx+C

F(pi/3)=1/6pi+1/4 sin (2/3pi)-tan^2(1/3pi)-ln (cos (1/3pi)-cos (1/3pi) +CF(π3)=16π+14sin(23π)−tan2(13π)−ln(cos(13π)−cos(13π)+C

=1/6pi+1/8sqrt 3 - 3-ln(1/2)-1/2+C=16π+18√3−3−ln(12)−12+C

=1=1

So, C=9/2-1/8sqrt 3- ln 2-pi/6C=92−18√3−ln2−π6

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