Find the integrals of the functions ${sin}^{3} (2 x + 1)$

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Solution

Let $I = \int {sin}^{3} (2 x + 1)$
$\Rightarrow \int {sin}^{3} (2 x + 1) d x = \int {sin}^{2} (2 x + 1) \cdot sin (2 x + 1) d x$
$= \int (1 - {cos}^{2} (2 x + 1)) sin (2 x + 1) d x$
Put $cos (2 x + 1) = t$
$\Rightarrow - 2 sin (2 x + 1) d x = d t$
$\Rightarrow sin (2 x + 1) d x = \frac{- d t}{2}$
$\Rightarrow I = \frac{- 1}{2} \int (1 - t^{2}) d t$
$= \frac{- 1}{2} {t - \frac{t^{3}}{3}}$
$= \frac{- 1}{2} {cos (2 x + 1) - \frac{{cos}^{3} (2 x + 1)}{3}}$
$= \frac{- cos (2 x + 1)}{2} + \frac{{cos}^{3} (2 x + 1)}{6} + C$

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