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