Find the integrals of the functions ${sin}^{2} (2 x + 5)$

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Solution

${sin}^{2} (2 x + 5) = \frac{1 - cos 2 (2 x + 5)}{2} = \frac{1 - cos (4 x + 10)}{2}$
$\Rightarrow \int {sin}^{2} (2 x + 5) d x = \int \frac{1 - cos (4 x + 10)}{2} d x$
$= \frac{1}{2} \int 1 d x - \frac{1}{2} \int cos (4 x + 10) d x$
$= \frac{1}{2} x - \frac{1}{2} (\frac{sin (4 x + 10)}{4}) + C$
$= \frac{1}{2} x - \frac{1}{8} sin (4 x + 10) + C$

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