What is the integral of (x^2+4x+9)cos(3x) ?

The integral of (x^2+4x+9)cos(3x) is 1/27 (9x^2sin(3x)+6xcos(3x)+36xsin(3x)+79sin(3x)+12cos(3x))+C

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integral of (x^2+4x+9)cos(3x)

\int (x^{2} + 4 x + 9) cos (3 x) d x

\frac{1}{27} (9 x^{2} sin (3 x) + 6 x cos (3 x) + 36 x sin (3 x) + 79 sin (3 x) + 12 cos (3 x)) + C

Solution steps

\int (x^{2} + 4 x + 9) cos (3 x) d x

Apply u-substitution

: \int \frac{1}{3} cos (u) (\frac{u ^{2}}{9} + \frac{4 u}{3} + 9) d u

= \int \frac{1}{3} cos (u) (\frac{u ^{2}}{9} + \frac{4 u}{3} + 9) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int cos (u) (\frac{u ^{2}}{9} + \frac{4 u}{3} + 9) d u

Expand

cos (u) (\frac{u ^{2}}{9} + \frac{4 u}{3} + 9) : \frac{u ^{2} cos ( u )}{9} + \frac{4 u cos ( u )}{3} + 9 cos (u)

= \frac{1}{3} \cdot \int \frac{u ^{2} cos ( u )}{9} + \frac{4 u cos ( u )}{3} + 9 cos (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int \frac{u ^{2} cos ( u )}{9} d u + \int \frac{4 u cos ( u )}{3} d u + \int 9 cos (u) d u)

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

\int \frac{4 u cos ( u )}{3} d u = \frac{4}{3} (u sin (u) + cos (u))

\int 9 cos (u) d u = 9 sin (u)

= \frac{1}{3} (\frac{1}{9} (u^{2} sin (u) - 2 (- u cos (u) + sin (u))) + \frac{4}{3} (u sin (u) + cos (u)) + 9 sin (u))

Substitute back

u = 3 x

= \frac{1}{3} (\frac{1}{9} ((3 x)^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x))) + \frac{4}{3} (3 x sin (3 x) + cos (3 x)) + 9 sin (3 x))

Simplify

\frac{1}{3} (\frac{1}{9} ((3 x)^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x))) + \frac{4}{3} (3 x sin (3 x) + cos (3 x)) + 9 sin (3 x)) : \frac{1}{27} (9 x^{2} sin (3 x) + 6 x cos (3 x) + 36 x sin (3 x) + 79 sin (3 x) + 12 cos (3 x))

= \frac{1}{27} (9 x^{2} sin (3 x) + 6 x cos (3 x) + 36 x sin (3 x) + 79 sin (3 x) + 12 cos (3 x))

Add a constant to the solution

= \frac{1}{27} (9 x^{2} sin (3 x) + 6 x cos (3 x) + 36 x sin (3 x) + 79 sin (3 x) + 12 cos (3 x)) + C

\int \frac{1}{x ^{2} + 4 x + 10} d x

\int 2 x - x d x

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

\int \frac{20 x}{x ^{2} + 1} d x

\int \frac{1}{25 + 9 y ^{2}} d y

What is the integral of (x^2+4x+9)cos(3x) ?
The integral of (x^2+4x+9)cos(3x) is 1/27 (9x^2sin(3x)+6xcos(3x)+36xsin(3x)+79sin(3x)+12cos(3x))+C