What is the integral of sin(-4x) ?

The integral of sin(-4x) is 1/4 cos(4x)+C

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integral of sin(-4x)

\int sin (- 4 x) d x

\frac{1}{4} cos (4 x) + C

Solution steps

\int sin (- 4 x) d x

Apply u-substitution

: \int - \frac{1}{4} sin (u) d u

= \int - \frac{1}{4} sin (u) d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int sin (u) d u

Use the common integral:

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

= - \frac{1}{4} (- cos (u))

Substitute back

u = - 4 x

= - \frac{1}{4} (- cos (- 4 x))

Simplify

- \frac{1}{4} (- cos (- 4 x)) : \frac{1}{4} cos (4 x)

= \frac{1}{4} cos (4 x)

Add a constant to the solution

= \frac{1}{4} cos (4 x) + C

\int 6 x + 3 d x

\int 1 - x d x

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\int sec^{4} (x t a) n^{5} x d x

n = 1 \sum \infty \frac{4}{n ( n + 7 )}