How do you find the integral of #cos^2(2x)dx#?

1 Answer
Aug 28, 2016

#\frac{1}{2}\(x+\frac{1}{4}\sin \(4x))+C#

Explanation:

#int cos ^2(2x)dx#

using the following identity,

#cos ^2(x)=frac{1+cos (2x)}{2}#

#=int frac{1+cos (2cdot 2x)}{2}dx#

taking the constant out,

#int acdot f(x)dx=acdot int f(x)dx#

so,#=frac{1}{2}int 1+cos (2cdot 2x)dx#

applying the sum rule,

#int f(x)pm g(x)dx=int f(x)dxpm int g(x)dx#

we have,#int 1dx# #=x#
and,
#int cos (2cdot 2x)dx=frac{1}{4}sin (4x)#

finally,
#=frac{1}{2}(x+frac{1}{4}sin (4x))#

adding constant,we get,
#frac{1}{2}(x+frac{1}{4}sin (4x))+C#