Can we calculate integral of cos^4x as (cos^2x)^2 instead of cos^3x*cosx?

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Answer 1 · 2018-06-13T12:25:04

#int cos^4xdx = (3x)/8 +sin(2x)/4 + sin(4x)/32 + C#

Use the trigonometric identity:

#cos^2 alpha = (1+cos(2alpha))/2#

Then:

#int cos^4xdx = int (cos^2x)^2dx#

#int cos^4xdx = int (1+cos(2x))^2/4dx#

using the linearity of the integral:

#int cos^4xdx = 1/4 int dx +1/2 int cos(2x)dx +1/4 int cos^2(2x)dx#

#int cos^4xdx = x/4 +sin(2x)/4 +1/4 int (1+cos(4x))/2dx#

#int cos^4xdx = x/4 +sin(2x)/4 +1/8 int dx + 1/8 intcos(4x)dx#

#int cos^4xdx = (3x)/8 +sin(2x)/4 + sin(4x)/32 + C#