Mathematics

Integrate: $$\sin ^ { 3 } ( 2 x + 1 )$$


SOLUTION
$$I=\displaystyle \int \sin^3 (2x + 1) dx $$

Let, $$2x + 1 = t$$

$$2 \, dx = dt$$

$$I=\dfrac{1}{2} \displaystyle \int \sin^3 (t) dt $$

We know that,
$$\sin (3t) = 3 \sin t - 4 \sin^3 t$$

$$4 \sin^3 t = 3 \sin t - \sin (3t)$$

$$\sin^3 t = \dfrac{1}{4} [3 \sin t - \sin (3t)]$$

$$I=\dfrac{1}{2} \left[\displaystyle \int \dfrac{1}{4} (3 \sin t \, dt) - \int \dfrac{1}{4} \sin (3t) dt \right]$$

$$I=\dfrac{1}{2} \times \dfrac{1}{4} \left[3 \cos t - \dfrac{\cos 3t}{3} \right] + c$$. substitute $$t = 2x + 1$$

$$I=\dfrac{1}{24} [ 9 \cos (2x + 1) - \cos (6x + 3) ] + c$$.
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Subjective Medium Published on 17th 09, 2020
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