What is the integral of (x^2+5)cos(3x) ?

The integral of (x^2+5)cos(3x) is 1/3 x^2sin(3x)+2/9 xcos(3x)+43/27 sin(3x)+C

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integral of (x^2+5)cos(3x)

\int (x^{2} + 5) cos (3 x) d x

\frac{1}{3} x^{2} sin (3 x) + \frac{2}{9} x cos (3 x) + \frac{43}{27} sin (3 x) + C

Solution steps

\int (x^{2} + 5) cos (3 x) d x

Expand

(x^{2} + 5) cos (3 x) : x^{2} cos (3 x) + 5 cos (3 x)

= \int x^{2} cos (3 x) + 5 cos (3 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^{2} cos (3 x) d x + \int 5 cos (3 x) d x

\int x^{2} cos (3 x) d x = \frac{1}{27} (9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x)))

\int 5 cos (3 x) d x = \frac{5}{3} sin (3 x)

= \frac{1}{27} (9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x))) + \frac{5}{3} sin (3 x)

Simplify

\frac{1}{27} (9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x))) + \frac{5}{3} sin (3 x) : \frac{1}{3} x^{2} sin (3 x) + \frac{2}{9} x cos (3 x) + \frac{43}{27} sin (3 x)

= \frac{1}{3} x^{2} sin (3 x) + \frac{2}{9} x cos (3 x) + \frac{43}{27} sin (3 x)

Add a constant to the solution

= \frac{1}{3} x^{2} sin (3 x) + \frac{2}{9} x cos (3 x) + \frac{43}{27} sin (3 x) + C

f (x) = x^{3} cos (2 x)

x \frac{d y}{d x} = x^{3} + 9 x^{3} y

y^{^{''}} = 6 x, y (0) = 1, y^{^{'}} (0) = - 1

\int \frac{1}{49 - 64 x ^{2}} d x

\frac{\partial}{\partial y} (x^{4} + x^{2} y + x + 5)

What is the integral of (x^2+5)cos(3x) ?
The integral of (x^2+5)cos(3x) is 1/3 x^2sin(3x)+2/9 xcos(3x)+43/27 sin(3x)+C