What is the integral of 6xcos(3x) ?

The integral of 6xcos(3x) is 2xsin(3x)+2/3 cos(3x)+C

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integral of 6xcos(3x)

\int 6 x cos (3 x) d x

2 x sin (3 x) + \frac{2}{3} cos (3 x) + C

Solution steps

\int 6 x cos (3 x) d x

Take the constant out:

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

= 6 \cdot \int x cos (3 x) d x

Apply u-substitution

= 6 \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

= 6 \cdot \frac{1}{9} \cdot \int u cos (u) d u

Apply Integration By Parts

= 6 \cdot \frac{1}{9} (u sin (u) - \int sin (u) d u)

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

= 6 \cdot \frac{1}{9} (u sin (u) - (- cos (u)))

Substitute back

u = 3 x

= 6 \cdot \frac{1}{9} (3 x sin (3 x) - (- cos (3 x)))

Simplify

6 \cdot \frac{1}{9} (3 x sin (3 x) - (- cos (3 x))) : 2 x sin (3 x) + \frac{2}{3} cos (3 x)

= 2 x sin (3 x) + \frac{2}{3} cos (3 x)

Add a constant to the solution

= 2 x sin (3 x) + \frac{2}{3} cos (3 x) + C

What is the integral of 6xcos(3x) ?
The integral of 6xcos(3x) is 2xsin(3x)+2/3 cos(3x)+C