What is the integral of (4x^2-3x+1)cos(x) ?

The integral of (4x^2-3x+1)cos(x) is 4x^2sin(x)+8xcos(x)-3xsin(x)-7sin(x)-3cos(x)+C

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integral of (4x^2-3x+1)cos(x)

\int (4 x^{2} - 3 x + 1) cos (x) d x

4 x^{2} sin (x) + 8 x cos (x) - 3 x sin (x) - 7 sin (x) - 3 cos (x) + C

Solution steps

\int (4 x^{2} - 3 x + 1) cos (x) d x

Expand

(4 x^{2} - 3 x + 1) cos (x) : 4 x^{2} cos (x) - 3 x cos (x) + cos (x)

= \int 4 x^{2} cos (x) - 3 x cos (x) + cos (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} cos (x) d x - \int 3 x cos (x) d x + \int cos (x) d x

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

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

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

= 4 (x^{2} sin (x) - 2 (- x cos (x) + sin (x))) - 3 (x sin (x) + cos (x)) + sin (x)

Simplify

4 (x^{2} sin (x) - 2 (- x cos (x) + sin (x))) - 3 (x sin (x) + cos (x)) + sin (x) : 4 x^{2} sin (x) + 8 x cos (x) - 3 x sin (x) - 7 sin (x) - 3 cos (x)

= 4 x^{2} sin (x) + 8 x cos (x) - 3 x sin (x) - 7 sin (x) - 3 cos (x)

Add a constant to the solution

= 4 x^{2} sin (x) + 8 x cos (x) - 3 x sin (x) - 7 sin (x) - 3 cos (x) + C

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What is the integral of (4x^2-3x+1)cos(x) ?
The integral of (4x^2-3x+1)cos(x) is 4x^2sin(x)+8xcos(x)-3xsin(x)-7sin(x)-3cos(x)+C