compute the integral: #int cos^2 x dx # ?

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Answer 1 · 2018-04-02T07:59:19

#int cos^2 x dx = 1/4 sin2x + 1/2x + c #

Use trig identities...

#cos^2x - sin^2 x = cos2x #

#=> cos^2 x - ( 1- sin^2 x ) #

#=> 2cos^2 x -1 #

#=> 2cos^2 x -1 = cos2x #

#=> cos^2 x = 1/2 (cos2x + 1 )#

#=> 1/2 int cos2x +1 dx #

#=> 1/2 ( 1/2sin2x +x + c_0 )#

#=> 1/4 sin2x + 1/2 x + c_1 #

#int cos^2 x dx = 1/4 sin2x + 1/2x + c #