What is the integral of sin(3x)2x ?

The integral of sin(3x)2x is 2(-1/3 xcos(3x)+1/9 sin(3x))+C

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integral of sin(3x)2x

\int sin (3 x) 2 x d x

2 (- \frac{1}{3} x cos (3 x) + \frac{1}{9} sin (3 x)) + C

Solution steps

\int sin (3 x) \cdot 2 x d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int sin (3 x) x d x

Apply Integration By Parts

: - \frac{1}{3} x cos (3 x) - \int - \frac{1}{3} cos (3 x) d x

= 2 (- \frac{1}{3} x cos (3 x) - \int - \frac{1}{3} cos (3 x) d x)

\int - \frac{1}{3} cos (3 x) d x = - \frac{1}{9} sin (3 x)

= 2 (- \frac{1}{3} x cos (3 x) - (- \frac{1}{9} sin (3 x)))

Simplify

= 2 (- \frac{1}{3} x cos (3 x) + \frac{1}{9} sin (3 x))

Add a constant to the solution

= 2 (- \frac{1}{3} x cos (3 x) + \frac{1}{9} sin (3 x)) + C

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What is the integral of sin(3x)2x ?
The integral of sin(3x)2x is 2(-1/3 xcos(3x)+1/9 sin(3x))+C