What is the integral of-2xcos(6x) ?

The integral of-2xcos(6x) is -1/18 (6xsin(6x)+cos(6x))+C

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integral of-2xcos(6x)

\int - 2 x cos (6 x) d x

- \frac{1}{18} (6 x sin (6 x) + cos (6 x)) + C

Solution steps

\int - 2 x cos (6 x) d x

Take the constant out:

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

= - 2 \cdot \int x cos (6 x) d x

Apply u-substitution

: \int \frac{u cos ( u )}{36} d u

= - 2 \cdot \int \frac{u cos ( u )}{36} d u

Take the constant out:

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

= - 2 \cdot \frac{1}{36} \cdot \int u cos (u) d u

Apply Integration By Parts

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

= - 2 \cdot \frac{1}{36} (u sin (u) - \int sin (u) d u)

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

= - 2 \cdot \frac{1}{36} (u sin (u) - (- cos (u)))

Substitute back

u = 6 x

= - 2 \cdot \frac{1}{36} (6 x sin (6 x) - (- cos (6 x)))

Simplify

- 2 \cdot \frac{1}{36} (6 x sin (6 x) - (- cos (6 x))) : - \frac{1}{18} (6 x sin (6 x) + cos (6 x))

= - \frac{1}{18} (6 x sin (6 x) + cos (6 x))

Add a constant to the solution

= - \frac{1}{18} (6 x sin (6 x) + cos (6 x)) + C

\frac{\partial}{\partial x} (ln (\frac{x y}{y + z}))

\frac{6}{s ( s ^{2} + 2 s + 6 )}

f (x) = \frac{1 + x ^{2}}{x ^{2} - 1}

y = ln (x + x^{2} - 6)

\frac{d}{d a} (csc (a) tan (a) cos (a) - csc^{2} (a))

What is the integral of-2xcos(6x) ?
The integral of-2xcos(6x) is -1/18 (6xsin(6x)+cos(6x))+C