What is the integral from 0 to 2pi of sin(x)sin(2x) ?

The integral from 0 to 2pi of sin(x)sin(2x) is 0

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integral from 0 to 2pi of sin(x)sin(2x)

\int_{0}^{2 π} sin (x) sin (2 x) d x

0

Solution steps

\int_{0}^{2 π} sin (x) sin (2 x) d x

Rewrite using trig identities

= \int_{0}^{2 π} \frac{1}{2} (- cos (3 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_{0}^{2 π} - cos (3 x) + cos (- x) d x

Simplify

- cos (3 x) + cos (- x) : cos (x) - cos (3 x)

= \frac{1}{2} \cdot \int_{0}^{2 π} cos (x) - cos (3 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}^{2 π} cos (x) d x - \int_{0}^{2 π} cos (3 x) d x)

\int_{0}^{2 π} cos (x) d x = 0

\int_{0}^{2 π} cos (3 x) d x = 0

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

Simplify

= 0

What is the integral from 0 to 2pi of sin(x)sin(2x) ?
The integral from 0 to 2pi of sin(x)sin(2x) is 0