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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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