What is the integral of (3-2x)sin(5-4x) ?

The integral of (3-2x)sin(5-4x) is 3/4 cos(5-4x)-2(1/4 xcos(5-4x)+1/16 sin(5-4x))+C

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integral of (3-2x)sin(5-4x)

\int (3 - 2 x) sin (5 - 4 x) d x

\frac{3}{4} cos (5 - 4 x) - 2 (\frac{1}{4} x cos (5 - 4 x) + \frac{1}{16} sin (5 - 4 x)) + C

Solution steps

\int (3 - 2 x) sin (5 - 4 x) d x

Expand

(3 - 2 x) sin (5 - 4 x) : 3 sin (5 - 4 x) - 2 x sin (5 - 4 x)

= \int 3 sin (5 - 4 x) - 2 x sin (5 - 4 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 3 sin (5 - 4 x) d x - \int 2 x sin (5 - 4 x) d x

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

\int 2 x sin (5 - 4 x) d x = 2 (\frac{1}{4} x cos (5 - 4 x) + \frac{1}{16} sin (5 - 4 x))

= \frac{3}{4} cos (5 - 4 x) - 2 (\frac{1}{4} x cos (5 - 4 x) + \frac{1}{16} sin (5 - 4 x))

Add a constant to the solution

= \frac{3}{4} cos (5 - 4 x) - 2 (\frac{1}{4} x cos (5 - 4 x) + \frac{1}{16} sin (5 - 4 x)) + C

3^{x}

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\int (x^{3} - 3 x + 2) d x

What is the integral of (3-2x)sin(5-4x) ?
The integral of (3-2x)sin(5-4x) is 3/4 cos(5-4x)-2(1/4 xcos(5-4x)+1/16 sin(5-4x))+C