What is the integral of cos(3-6x) ?

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

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integral of cos(3-6x)

\int cos (3 - 6 x) d x

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

Solution steps

\int cos (3 - 6 x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = 3 - 6 x

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

Add a constant to the solution

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

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