What is the integral of sin(9x)cos(9x) ?

The integral of sin(9x)cos(9x) is -1/18 cos^2(9x)+C

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integral of sin(9x)cos(9x)

\int sin (9 x) cos (9 x) d x

- \frac{1}{18} cos^{2} (9 x) + C

Solution steps

\int sin (9 x) cos (9 x) d x

Apply u-substitution

: \int - \frac{u}{9} d u

= \int - \frac{u}{9} d u

Take the constant out:

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

= - \frac{1}{9} \cdot \int u d u

Apply the Power Rule

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

= - \frac{1}{9} \cdot \frac{u ^{2}}{2}

Substitute back

u = cos (9 x)

= - \frac{1}{9} \cdot \frac{cos ^{2} ( 9 x )}{2}

Simplify

- \frac{1}{9} \cdot \frac{cos ^{2} ( 9 x )}{2} : - \frac{1}{18} cos^{2} (9 x)

= - \frac{1}{18} cos^{2} (9 x)

Add a constant to the solution

= - \frac{1}{18} cos^{2} (9 x) + C

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What is the integral of sin(9x)cos(9x) ?
The integral of sin(9x)cos(9x) is -1/18 cos^2(9x)+C