What is the integral of (x^3-3x+8)cos(x) ?

The integral of (x^3-3x+8)cos(x) is x^3sin(x)+3x^2cos(x)-9xsin(x)-9cos(x)+8sin(x)+C

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integral of (x^3-3x+8)cos(x)

\int (x^{3} - 3 x + 8) cos (x) d x

x^{3} sin (x) + 3 x^{2} cos (x) - 9 x sin (x) - 9 cos (x) + 8 sin (x) + C

Solution steps

\int (x^{3} - 3 x + 8) cos (x) d x

Expand

(x^{3} - 3 x + 8) cos (x) : x^{3} cos (x) - 3 x cos (x) + 8 cos (x)

= \int x^{3} cos (x) - 3 x cos (x) + 8 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 x^{3} cos (x) d x - \int 3 x cos (x) d x + \int 8 cos (x) d x

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

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

\int 8 cos (x) d x = 8 sin (x)

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

Simplify

x^{3} sin (x) - 3 (- x^{2} cos (x) + 2 (x sin (x) + cos (x))) - 3 (x sin (x) + cos (x)) + 8 sin (x) : x^{3} sin (x) + 3 x^{2} cos (x) - 9 x sin (x) - 9 cos (x) + 8 sin (x)

= x^{3} sin (x) + 3 x^{2} cos (x) - 9 x sin (x) - 9 cos (x) + 8 sin (x)

Add a constant to the solution

= x^{3} sin (x) + 3 x^{2} cos (x) - 9 x sin (x) - 9 cos (x) + 8 sin (x) + C

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

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

\int \frac{1}{y - 1} d y

\int tan^{2} (x) + cot^{2} (x) d x

\int \frac{3 x ^{2}}{cos ^{2} ( x ^{3} )} d x

What is the integral of (x^3-3x+8)cos(x) ?
The integral of (x^3-3x+8)cos(x) is x^3sin(x)+3x^2cos(x)-9xsin(x)-9cos(x)+8sin(x)+C