Mathematics

# Solve$\int \sin^{3}(2x-1) \ dx$

##### SOLUTION
$\sin^{3}(2x+1)dx$

$=\displaystyle\int \dfrac{3\sin (2x+1)-\sin (6x+3)}{4}dx$
$\left\{\sin 3x=3\sin x-4\sin^{3}x\right\}$

$=\dfrac{-3}{4}\dfrac{\cos (2x+1)}{2}+\dfrac{1}{4}\dfrac{\cos (6x+3)}{6}$

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

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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