Mathematics

# $\int { \sin ^{ 3 }{ \left( 2x+1 \right) } } dx$

##### SOLUTION
$\displaystyle\int{{\sin}^{3}{\left(2x+1\right)}dx}$
We know that $\sin{3\theta}=3\sin{\theta}-4{\sin}^{3}{\theta}$

$\Rightarrow\,{\sin}^{3}{\theta}=\dfrac{1}{4}\left[3\sin{\theta}-\sin{3\theta}\right]$

$=\dfrac{1}{4}\displaystyle\int{\left[3\sin{\left(2x+1\right)}-\sin{3\left(2x+1\right)}\right]dx}$

Let $t=2x+1\Rightarrow\,dt=2\,dx$

$=\dfrac{1}{8}\displaystyle\int{\left[3\sin{t}-\sin{3t}\right]dt}$

$=\dfrac{3}{8}\displaystyle\int{\sin{t}\,dt}-\dfrac{1}{8}\displaystyle\int{\sin{3t}dt}$

$=\dfrac{-3}{8}\cos{t}+\dfrac{1}{24}\cos{3t}+c$

$=\dfrac{-3}{8}\cos{\left(2x+1\right)}+\dfrac{1}{24}\cos{3\left(2x+1\right)}+c$

$=\dfrac{-3}{8}\cos{\left(2x+1\right)}+\dfrac{1}{24}\cos{\left(6x+3\right)}+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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