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

The integral of-2cos(6x) is -1/3 sin(6x)+C

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integral of-2cos(6x)

\int - 2 cos (6 x) d x

- \frac{1}{3} sin (6 x) + C

Solution steps

\int - 2 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 cos (6 x) d x

Apply u-substitution

: \int cos (u) \frac{1}{6} d u

= - 2 \cdot \int cos (u) \frac{1}{6} d u

Take the constant out:

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

= - 2 \cdot \frac{1}{6} \cdot \int cos (u) d u

Use the common integral:

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

= - 2 \cdot \frac{1}{6} sin (u)

Substitute back

u = 6 x

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

Simplify

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

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

Add a constant to the solution

= - \frac{1}{3} sin (6 x) + C

f (x) = 8 x^{15} - 6 x^{3} + 1

\frac{d y}{d t} = y^{2} - y

x^{3} - 10 x, 15 x

y^{^{'''}} + y^{^{''}} - 4 y^{^{'}} - 4 y = 0

\int_{0}^{6} (6 x + 2) d x