What is the integral from 0 to pi/2 of cos(9x)cos(18x) ?

The integral from 0 to pi/2 of cos(9x)cos(18x) is 1/27

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integral from 0 to pi/2 of cos(9x)cos(18x)

\int_{0}^{\frac{π}{2}} cos (9 x) cos (18 x) d x

\frac{1}{27}

Decimal Notation

0.03703 \dots

Solution steps

\int_{0}^{\frac{π}{2}} cos (9 x) cos (18 x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{2}} \frac{1}{2} (cos (27 x) + cos (- 9 x)) 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_{0}^{\frac{π}{2}} cos (27 x) + cos (- 9 x) d x

Simplify

cos (27 x) + cos (- 9 x) : cos (9 x) + cos (27 x)

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} cos (9 x) + cos (27 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{0}^{\frac{π}{2}} cos (9 x) d x + \int_{0}^{\frac{π}{2}} cos (27 x) d x)

\int_{0}^{\frac{π}{2}} cos (9 x) d x = \frac{1}{9}

\int_{0}^{\frac{π}{2}} cos (27 x) d x = - \frac{1}{27}

= \frac{1}{2} (\frac{1}{9} - \frac{1}{27})

Simplify

\frac{1}{2} (\frac{1}{9} - \frac{1}{27}) : \frac{1}{27}

= \frac{1}{27}

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What is the integral from 0 to pi/2 of cos(9x)cos(18x) ?
The integral from 0 to pi/2 of cos(9x)cos(18x) is 1/27