What is the integral from 0 to 2pi of sin(3x)cos(4x) ?

The integral from 0 to 2pi of sin(3x)cos(4x) is 0

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integral from 0 to 2pi of sin(3x)cos(4x)

\int_{0}^{2 π} sin (3 x) cos (4 x) d x

0

Solution steps

\int_{0}^{2 π} sin (3 x) cos (4 x) d x

Use the following identity:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

= \int_{0}^{2 π} \frac{sin ( 3 x + 4 x ) + sin ( 3 x - 4 x )}{2} 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} sin (3 x + 4 x) + sin (3 x - 4 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} sin (3 x + 4 x) d x + \int_{0}^{2 π} sin (3 x - 4 x) d x)

\int_{0}^{2 π} sin (3 x + 4 x) d x = 0

\int_{0}^{2 π} sin (3 x - 4 x) d x = 0

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

Simplify

= 0

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What is the integral from 0 to 2pi of sin(3x)cos(4x) ?
The integral from 0 to 2pi of sin(3x)cos(4x) is 0