What is the integral from 0 to pi of sin(9x)sin(7x) ?

The integral from 0 to pi of sin(9x)sin(7x) is 0

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integral from 0 to pi of sin(9x)sin(7x)

\int_{0}^{π} sin (9 x) sin (7 x) d x

0

Solution steps

\int_{0}^{π} sin (9 x) sin (7 x) d x

Rewrite using trig identities

= \int_{0}^{π} \frac{1}{2} (- cos (16 x) + cos (2 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}^{π} - cos (16 x) + cos (2 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}^{π} cos (16 x) d x + \int_{0}^{π} cos (2 x) d x)

\int_{0}^{π} cos (16 x) d x = 0

\int_{0}^{π} cos (2 x) d x = 0

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

Simplify

= 0

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What is the integral from 0 to pi of sin(9x)sin(7x) ?
The integral from 0 to pi of sin(9x)sin(7x) is 0