sin^3(2x+1)dx =∫3sin (2x+1) sin (6x+3)4dx {sin 3x=3sin x 4sin^3x} = 34cos (2x+1)2+14cos (6x+3)6 = 38cos (2x+1)+124cos (6x+3)+C ...

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Integration of Trigonometric Functions

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Question

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

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Solution

${sin}^{3} (2 x + 1) d x$

$= \int \frac{3 sin (2 x + 1) - sin (6 x + 3)}{4} d x$
${sin 3 x = 3 sin x - 4 {sin}^{3} x}$

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

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

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