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

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

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integral of (2x+3)sin(3x)

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

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

Solution steps

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

Expand

(2 x + 3) sin (3 x) : 2 x sin (3 x) + 3 sin (3 x)

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 3 sin (3 x) d x = - cos (3 x)

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

Add a constant to the solution

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

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