What is the integral of (x^2+9x)cos(x) ?

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

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integral of (x^2+9x)cos(x)

\int (x^{2} + 9 x) cos (x) d x

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

Solution steps

\int (x^{2} + 9 x) cos (x) d x

Apply Integration By Parts

= (x^{2} + 9 x) sin (x) - \int (2 x + 9) sin (x) d x

\int (2 x + 9) sin (x) d x = - cos (x) (2 x + 9) + 2 sin (x)

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

Simplify

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

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

Add a constant to the solution

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

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