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
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