integrate sin(3x) — step-by-step solution

Let's compute the integral:

\int \sin (3 x) d x

To use the known integral of

\sin (u)

, let

u = 3 x

. Then

du = 3 dx

, so

dx = \frac{1}{3} \cdot du

We apply

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

, then substitute

u = 3 x

back

ⓘ The derivative of

- \frac{\cos (3 x)}{3}

\sin (3 x)

, so

- \frac{\cos (3 x)}{3}

is an antiderivative of

\sin (3 x)

\int \sin (3 x) d x

- \frac{\cos (3 x)}{3}

- \frac{\cos (3 x)}{3}

▶An indefinite integral is defined up to an additive constant C; we therefore obtain:

- \frac{\cos (3 x)}{3} + C