What is the integral of (3x-2)sin(1-x) ?

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

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integral of (3x-2)sin(1-x)

\int (3 x - 2) sin (1 - x) d x

- 3 (- x cos (1 - x) - sin (1 - x)) - 2 cos (1 - x) + C

Solution steps

\int (3 x - 2) sin (1 - x) d x

Expand

(3 x - 2) sin (1 - x) : 3 x sin (1 - x) - 2 sin (1 - x)

= \int 3 x sin (1 - x) - 2 sin (1 - 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 x sin (1 - x) d x - \int 2 sin (1 - x) d x

\int 3 x sin (1 - x) d x = - 3 (- x cos (1 - x) - sin (1 - x))

\int 2 sin (1 - x) d x = 2 cos (1 - x)

= - 3 (- x cos (1 - x) - sin (1 - x)) - 2 cos (1 - x)

Add a constant to the solution

= - 3 (- x cos (1 - x) - sin (1 - x)) - 2 cos (1 - x) + C

\int \frac{3 x}{2 x ^{2} - 6} d x

\int \frac{1}{( 3 x + 4 ) ^{3}} d x

\int x x^{2} - 25 d x

\int x^{2} + 121 d x

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