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

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Answer 1 · 2016-08-28T09:46:55

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

$\int {cos}^{2} (2 x) d x$ $int cos ^2(2x)dx$

using the following identity,

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

$= \int \frac{1 + cos (2 \cdot 2 x)}{2} d x$ $=int frac{1+cos (2cdot 2x)}{2}dx$

taking the constant out,

$\int a \cdot f (x) d x = a \cdot \int f (x) d x$ $int acdot f(x)dx=acdot int f(x)dx$

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

applying the sum rule,

$\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x$ $int f(x)pm g(x)dx=int f(x)dxpm int g(x)dx$

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

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

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

How do you find the integral of cos2(2x)dxcos^2(2x)dx?