What is the integral of (7-5x)sin(3-2x) ?

The integral of (7-5x)sin(3-2x) is 7/2 cos(3-2x)-5(1/2 xcos(3-2x)+1/4 sin(3-2x))+C

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integral of (7-5x)sin(3-2x)

\int (7 - 5 x) sin (3 - 2 x) d x

\frac{7}{2} cos (3 - 2 x) - 5 (\frac{1}{2} x cos (3 - 2 x) + \frac{1}{4} sin (3 - 2 x)) + C

Solution steps

\int (7 - 5 x) sin (3 - 2 x) d x

Expand

(7 - 5 x) sin (3 - 2 x) : 7 sin (3 - 2 x) - 5 x sin (3 - 2 x)

= \int 7 sin (3 - 2 x) - 5 x sin (3 - 2 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 7 sin (3 - 2 x) d x - \int 5 x sin (3 - 2 x) d x

\int 7 sin (3 - 2 x) d x = \frac{7}{2} cos (3 - 2 x)

\int 5 x sin (3 - 2 x) d x = 5 (\frac{1}{2} x cos (3 - 2 x) + \frac{1}{4} sin (3 - 2 x))

= \frac{7}{2} cos (3 - 2 x) - 5 (\frac{1}{2} x cos (3 - 2 x) + \frac{1}{4} sin (3 - 2 x))

Add a constant to the solution

= \frac{7}{2} cos (3 - 2 x) - 5 (\frac{1}{2} x cos (3 - 2 x) + \frac{1}{4} sin (3 - 2 x)) + C

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What is the integral of (7-5x)sin(3-2x) ?
The integral of (7-5x)sin(3-2x) is 7/2 cos(3-2x)-5(1/2 xcos(3-2x)+1/4 sin(3-2x))+C