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

The integral from-pi to 0 of sin(3x)cos(2x) is -6/5

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

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

- \frac{6}{5}

Solution steps

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

Use the following identity:

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

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

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

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

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

Simplify

\frac{1}{2} (- \frac{2}{5} - 2) : - \frac{6}{5}

= - \frac{6}{5}

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What is the integral from-pi to 0 of sin(3x)cos(2x) ?
The integral from-pi to 0 of sin(3x)cos(2x) is -6/5