What is the integral of 9xsin(2-3x) ?

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

integral of 9xsin(2-3x)

\int 9 x sin (2 - 3 x) d x

3 x cos (2 - 3 x) + sin (2 - 3 x) + C

Solution steps

\int 9 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

= 9 \cdot \int x sin (2 - 3 x) d x

Apply Integration By Parts

= 9 (\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)

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

Simplify

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

= 3 x cos (2 - 3 x) + sin (2 - 3 x)

Add a constant to the solution

= 3 x cos (2 - 3 x) + sin (2 - 3 x) + C

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