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

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

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integral of 2sin(3x)cos(3x)

\int 2 sin (3 x) cos (3 x) d x

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

Solution steps

\int 2 sin (3 x) cos (3 x) d x

Take the constant out:

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

= 2 \cdot \int sin (3 x) cos (3 x) d x

Apply u-substitution

= 2 \cdot \int - \frac{u}{3} d u

Take the constant out:

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

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

Apply the Power Rule

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

Substitute back

u = cos (3 x)

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

Simplify

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

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

Add a constant to the solution

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

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