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

The integral of (2x-1)^2sin(x) is -4x^2cos(x)+8xsin(x)+4xcos(x)+7cos(x)-4sin(x)+C

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

\int (2 x - 1)^{2} sin (x) d x

- 4 x^{2} cos (x) + 8 x sin (x) + 4 x cos (x) + 7 cos (x) - 4 sin (x) + C

Solution steps

\int (2 x - 1)^{2} sin (x) d x

Expand

(2 x - 1)^{2} sin (x) : 4 x^{2} sin (x) - 4 x sin (x) + sin (x)

= \int 4 x^{2} sin (x) - 4 x sin (x) + sin (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 4 x^{2} sin (x) d x - \int 4 x sin (x) d x + \int sin (x) d x

\int 4 x^{2} sin (x) d x = 4 (- x^{2} cos (x) + 2 (x sin (x) + cos (x)))

\int 4 x sin (x) d x = 4 (- x cos (x) + sin (x))

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

= 4 (- x^{2} cos (x) + 2 (x sin (x) + cos (x))) - 4 (- x cos (x) + sin (x)) - cos (x)

Simplify

4 (- x^{2} cos (x) + 2 (x sin (x) + cos (x))) - 4 (- x cos (x) + sin (x)) - cos (x) : - 4 x^{2} cos (x) + 8 x sin (x) + 4 x cos (x) + 7 cos (x) - 4 sin (x)

= - 4 x^{2} cos (x) + 8 x sin (x) + 4 x cos (x) + 7 cos (x) - 4 sin (x)

Add a constant to the solution

= - 4 x^{2} cos (x) + 8 x sin (x) + 4 x cos (x) + 7 cos (x) - 4 sin (x) + C

What is the integral of (2x-1)^2sin(x) ?
The integral of (2x-1)^2sin(x) is -4x^2cos(x)+8xsin(x)+4xcos(x)+7cos(x)-4sin(x)+C