What is the integral from 0 to pi of sin(6x)cos(2x) ?

The integral from 0 to pi of sin(6x)cos(2x) is 0

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integral from 0 to pi of sin(6x)cos(2x)

\int_{0}^{π} sin (6 x) cos (2 x) d x

0

Solution steps

\int_{0}^{π} sin (6 x) cos (2 x) d x

Use the following identity:

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

= \int_{0}^{π} \frac{sin ( 6 x + 2 x ) + sin ( 6 x - 2 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}^{π} sin (6 x + 2 x) + sin (6 x - 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} sin (6 x + 2 x) d x + \int_{0}^{π} sin (6 x - 2 x) d x)

\int_{0}^{π} sin (6 x + 2 x) d x = 0

\int_{0}^{π} sin (6 x - 2 x) d x = 0

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

Simplify

= 0

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\int_{1}^{\frac{π}{2}} cos (x) d x

What is the integral from 0 to pi of sin(6x)cos(2x) ?
The integral from 0 to pi of sin(6x)cos(2x) is 0