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

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

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integral of cos(-6x)

\int cos (- 6 x) d x

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

Solution steps

\int cos (- 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 = - 6 x

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

Simplify

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

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

Add a constant to the solution

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

y = \frac{- 13}{4 x}

g (x) = x^{2} (1 - 3 x)

\int b x \cdot sin (\frac{π x}{2 a}) d x

\frac{\partial}{\partial y} (5 x^{2} - 2 x y + 3 y^{3})

(5.8) (3.9)