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

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

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integral of cos(-3x)

\int cos (- 3 x) d x

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

Solution steps

\int cos (- 3 x) d x

Apply u-substitution

: \int - \frac{1}{3} cos (u) d u

= \int - \frac{1}{3} cos (u) d u

Take the constant out:

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

= - \frac{1}{3} \cdot \int cos (u) d u

Use the common integral:

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

= - \frac{1}{3} sin (u)

Substitute back

u = - 3 x

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

Simplify

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

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

Add a constant to the solution

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

x y^{3} + 5 x y = 36, (6, 1)

\frac{\partial}{\partial x} (3 e x sin (y))

\int (x^{3} + 2) d x

\int - \frac{sin ^{2} ( x )}{cos ( x )} d x

\frac{d}{d x} (\frac{x ^{2} + 3}{x ^{2} - 1})