What is the integral of sin(24x) ?

The integral of sin(24x) is -1/24 cos(24x)+C

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integral of sin(24x)

\int sin (24 x) d x

- \frac{1}{24} cos (24 x) + C

Solution steps

\int sin (24 x) d x

Apply u-substitution

: \int sin (u) \frac{1}{24} d u

= \int sin (u) \frac{1}{24} d u

Take the constant out:

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

= \frac{1}{24} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{1}{24} (- cos (u))

Substitute back

u = 24 x

= \frac{1}{24} (- cos (24 x))

Simplify

= - \frac{1}{24} cos (24 x)

Add a constant to the solution

= - \frac{1}{24} cos (24 x) + C

\int_{0}^{2} 1 + x d x

\frac{x}{1 - \frac{1}{x ^{2}}}

\int 3 (3 x + 9)^{5} d x

(t^{2} + 2 t) \frac{d x}{d t} = 2 x + 7, x (1) = 1