What is the integral from-pi to pi of cos(2x)cos(3x) ?

The integral from-pi to pi of cos(2x)cos(3x) is 0

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integral from-pi to pi of cos(2x)cos(3x)

\int_{- π}^{π} cos (2 x) cos (3 x) d x

0

Solution steps

\int_{- π}^{π} cos (2 x) cos (3 x) d x

Rewrite using trig identities

= \int_{- π}^{π} \frac{1}{2} (cos (5 x) + cos (- 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_{- π}^{π} cos (5 x) + cos (- x) d x

Simplify

cos (5 x) + cos (- x) : cos (x) + cos (5 x)

= \frac{1}{2} \cdot \int_{- π}^{π} cos (x) + cos (5 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_{- π}^{π} cos (x) d x + \int_{- π}^{π} cos (5 x) d x)

\int_{- π}^{π} cos (x) d x = 0

\int_{- π}^{π} cos (5 x) d x = 0

= \frac{1}{2} (0 + 0)

Simplify

= 0

x \to 5 lim (\frac{2}{5 - x})

\frac{d}{d x} ((1 + \frac{8}{x})^{x})

\frac{d}{d x} (x + 1 - x)

\frac{d y}{d t} + e^{t} y = - 4 e^{t}

x \to 0 + lim (\frac{- 4 x ^{2} - 6 x}{x})

What is the integral from-pi to pi of cos(2x)cos(3x) ?
The integral from-pi to pi of cos(2x)cos(3x) is 0