How do you find the integral of #cos^3 2x dx#?

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Answer 1 · 2017-10-19T11:36:43

# int cos^3 2xdx=1/24(sin6x+9sin2x)+C.#

Recall that, #cos3theta=4cos^3theta-3costheta.#

#:. cos3theta+3costheta=4cos^3theta, or, #

# cos^3 theta=1/4(cos3theta+3costheta).#

Replacing #theta# by #2x,# we have,

#cos^3 2x=1/4(cos 6x+3cos 2x).#

#:. int cos^3 2xdx=int{1/4(cos 6x+3cos 2x)}dx,#

#=1/4intcos 6x dx+3/4intcos2xdx,#

#=1/4*sin(6x)/6+3/4*sin(2x)/2,#

#=1/24*sin6x+3/8*sin2x,#

# rArr int cos^3 2xdx=1/24(sin6x+9sin2x)+C.#

Answer 2 · 2017-10-19T17:53:31

see below

Another approach

#intcos^3 2xdx#

#=intcos2xcos^2 2xdx#

#=intcos2x(1-sin^2 2x)dx#

#=int(cos2x-sin^2 2xcos2x)dx#

integrating by inspection

#= 1/2sin2x-1/6sin^3 2x+C#

which can be shown by trig identities to be equivalent to the previous solution. This is left for the reader to verify