How do you find the integral of ${cos}^{3} 2 x d x$ $cos^3 2x dx$ ?

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Answer 1 · 2017-10-19T11:36:43

$\int {cos}^{3} 2 x d x = \frac{1}{24} (sin 6 x + 9 sin 2 x) + C .$ $int cos^3 2xdx=1/24(sin6x+9sin2x)+C.$

Recall that, $cos 3 θ = 4 {cos}^{3} θ - 3 cos θ .$ $cos3theta=4cos^3theta-3costheta.$

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

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

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

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