What is the integral of 6xsin(2-3x) ?

The integral of 6xsin(2-3x) is 6(1/3 xcos(2-3x)+1/9 sin(2-3x))+C

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integral of 6xsin(2-3x)

\int 6 x sin (2 - 3 x) d x

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

Solution steps

\int 6 x sin (2 - 3 x) d x

Take the constant out:

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

= 6 \cdot \int x sin (2 - 3 x) d x

Apply Integration By Parts

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

= 6 (\frac{1}{3} x cos (2 - 3 x) - \int \frac{1}{3} cos (2 - 3 x) d x)

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

= 6 (\frac{1}{3} x cos (2 - 3 x) - (- \frac{1}{9} sin (2 - 3 x)))

Simplify

= 6 (\frac{1}{3} x cos (2 - 3 x) + \frac{1}{9} sin (2 - 3 x))

Add a constant to the solution

= 6 (\frac{1}{3} x cos (2 - 3 x) + \frac{1}{9} sin (2 - 3 x)) + C

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What is the integral of 6xsin(2-3x) ?
The integral of 6xsin(2-3x) is 6(1/3 xcos(2-3x)+1/9 sin(2-3x))+C