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

1 Answer
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/6

Explanation:

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