What is the integral of ${cos}^{2} 6 x$ $cos^2 6x$ ?

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Answer 1 · 2016-04-11T15:56:00

$\frac{1}{2} (x + \frac{1}{12} sin 12 x)$ $1/2( x+1/12 sin 12x)$ +C

${cos}^{2} 6 x = \frac{1}{2} (1 + cos 12 x)$ $cos^2 6x = 1/2 (1+cos 12x)$ {identity $cos 2 x = 2 {cos}^{2} x - 1$ $cos2x= 2cos^2x -1$ }

$\int {cos}^{2} 6 x = \int \frac{1}{2} (1 + cos 12 x) d x$ $int cos^2 6x= int 1/2 (1+cos 12x) dx$

= $\frac{1}{2} (x + \frac{1}{12} sin 12 x)$ $1/2( x+1/12 sin 12x)$ +C

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