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

The integral from 0 to pi of sin(2x)cos(3x) is -4/5

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

\int_{0}^{π} sin (2 x) cos (3 x) d x

- \frac{4}{5}

Solution steps

\int_{0}^{π} sin (2 x) cos (3 x) d x

Use the following identity:

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

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

\int_{0}^{π} sin (2 x + 3 x) d x = \frac{2}{5}

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

= \frac{1}{2} (\frac{2}{5} - 2)

Simplify

\frac{1}{2} (\frac{2}{5} - 2) : - \frac{4}{5}

= - \frac{4}{5}

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What is the integral from 0 to pi of sin(2x)cos(3x) ?
The integral from 0 to pi of sin(2x)cos(3x) is -4/5