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

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Answer 1 · 2016-09-01T13:26:54

#f(x)= - 1/2 ( x + 1/2sin 2 x) -6+sqrt(3)/8+pi/6#

#f(x) = int -cos^2x dx#

# =- int cos^2x dx#

use identity #cos 2A = cos^2 A - sin^2 A = 2 cos^2 A - 1#

#implies - 1/2 int 1 + cos 2 x dx#

#f(x)= - 1/2 ( x + 1/2sin 2 x) + C#

#f(pi/3) = -6#

#implies -6 = - 1/2 ( pi/3 + 1/2sin (2pi )/3) + C#

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

#f(x)= - 1/2 ( x + 1/2sin 2 x) -6+sqrt(3)/8+pi/6#