Integral of square of Sine function

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$\int {\sin}^2\left(7x-1\right)\; dx$

**Answer**

$\int {\sin}^2\left(7x-1\right)\; dx$

Substitute $u=7x-1$. So, $du =7\; dx$

Or, $\frac{1}{7}\; du = dx$

And, $\int {\sin}^2\left(7x-1\right)\; dx = \frac{1}{7}\int {\sin}^2 u \;du$

$=\frac{1}{7}\left(\frac{1}{2}u-\frac{1}{4}\cos 2u \right) + C$

$=\frac{1}{14}u-\frac{1}{28}\cos 2u + C$

Substituting back $u=7x-1$

$=\frac{1}{14}\left(7x-1\right)-\frac{1}{28}\left( \cos 2\left(7x-1\right) \right) + C$

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