Find the integral of the function sin3 (2x + 1) - Vidyadip

Question

Find the integral of the function sin³ (2x + 1)

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Answer

Let $I=\int \sin ^{3}(2 x+1) d x$
$= \int \sin ^{3}(2 x+1) d x=\int \sin ^{2}(2 x+1) \cdot \sin (2 x+1) d x$
$= \int\left(1-\cos ^{2}(2 x+1)\right) \sin (2 x+1) d x$
Let cos (2x + 1) = t
$\Rightarrow$ -2sin(2x + 1)dx = dt
$\Rightarrow$ sin(2x + 1)dx = $\frac{-\mathrm{dt}}{2}$
$\therefore~ I=\frac{-1}{2} \int\left(1-t^{2}\right) d t$
$= \frac{-1}{2}\left\{t-\frac{t^{3}}{3}\right\}$
$= \frac{-1}{2}\left\{\cos (2 \mathrm{x}+1)-\frac{\cos ^{3}(2 \mathrm{x}+1)}{3}\right\}$
$= \frac{-\cos (2 \mathrm{x}+1)}{2}+\frac{\cos ^{3}(2 \mathrm{x}+1)}{6}+\mathrm{C}$

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