Integral of square of Cosine function

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$\int {\cos}^214x\; dx$

**Answer**

$\int {\cos}^214x\; dx$

Substitute $u=14x$. So, $du =14\; dx$

Or, $\frac{1}{14}\; du = dx$

And, $\int {\cos}^214x\; dx = \frac{1}{14}\int {\cos}^2 u \;du$

$=\frac{1}{14}\left(\frac{1}{2}u+\frac{1}{4}\cos 2u \right) + C$

$=\frac{1}{28}u+\frac{1}{56}\cos 2u + C$

Substituting back $u=14x$

$=\frac{1}{28}\left(14x\right)+\frac{1}{56}\left( \cos 2\left(14x\right) \right) + C$

$=\frac{1}{2}x+\frac{1}{56}\cos28x+ C$