What is the integral of 3xcos(3x) ?

The integral of 3xcos(3x) is xsin(3x)+1/3 cos(3x)+C

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integral of 3xcos(3x)

\int 3 x cos (3 x) d x

x sin (3 x) + \frac{1}{3} cos (3 x) + C

Solution steps

\int 3 x cos (3 x) d x

Take the constant out:

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

= 3 \cdot \int x cos (3 x) d x

Apply u-substitution

: \int \frac{u cos ( u )}{9} d u

= 3 \cdot \int \frac{u cos ( u )}{9} d u

Take the constant out:

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

= 3 \cdot \frac{1}{9} \cdot \int u cos (u) d u

Apply Integration By Parts

: u sin (u) - \int sin (u) d u

= 3 \cdot \frac{1}{9} (u sin (u) - \int sin (u) d u)

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

= 3 \cdot \frac{1}{9} (u sin (u) - (- cos (u)))

Substitute back

u = 3 x

= 3 \cdot \frac{1}{9} (3 x sin (3 x) - (- cos (3 x)))

Simplify

3 \cdot \frac{1}{9} (3 x sin (3 x) - (- cos (3 x))) : x sin (3 x) + \frac{1}{3} cos (3 x)

= x sin (3 x) + \frac{1}{3} cos (3 x)

Add a constant to the solution

= x sin (3 x) + \frac{1}{3} cos (3 x) + C

\frac{d}{d x} (4 sin (x) - 3 cos (x))

x \to 0 lim ((\frac{sin ( x )}{x})^{\frac{1}{x}})

y^{^{'}} - 2 y = 3 e^{2 t}

\frac{\partial}{\partial x} (8 x^{3} y - 7 y^{2} + 2 x)

f (x) = 2 cos (x), at x = \frac{π}{3}

What is the integral of 3xcos(3x) ?
The integral of 3xcos(3x) is xsin(3x)+1/3 cos(3x)+C