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

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

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integral of (sin(3x)cos(3x))

\int (sin (3 x) cos (3 x)) d x

- \frac{1}{6} cos^{2} (3 x) + C

Solution steps

\int sin (3 x) cos (3 x) d x

Apply u-substitution

: \int - \frac{u}{3} d u

= \int - \frac{u}{3} 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 u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= - \frac{1}{3} \cdot \frac{u ^{2}}{2}

Substitute back

u = cos (3 x)

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

Simplify

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

= - \frac{1}{6} cos^{2} (3 x)

Add a constant to the solution

= - \frac{1}{6} cos^{2} (3 x) + C

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