What is the integral of (sin(3x))/2 ?

The integral of (sin(3x))/2 is -1/6 cos(3x)+C

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integral of (sin(3x))/2

\int \frac{sin ( 3 x )}{2} d x

- \frac{1}{6} cos (3 x) + C

Solution steps

\int \frac{sin ( 3 x )}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int sin (3 x) d x

Apply u-substitution

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

= \frac{1}{2} \cdot \int sin (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{3} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{3} (- cos (u))

Substitute back

u = 3 x

= \frac{1}{2} \cdot \frac{1}{3} (- cos (3 x))

Simplify

\frac{1}{2} \cdot \frac{1}{3} (- cos (3 x)) : - \frac{1}{6} cos (3 x)

= - \frac{1}{6} cos (3 x)

Add a constant to the solution

= - \frac{1}{6} cos (3 x) + C

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

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(\frac{ln ( t )}{t})^{2}

What is the integral of (sin(3x))/2 ?
The integral of (sin(3x))/2 is -1/6 cos(3x)+C