What is the integral from 0 to pi of cos(6x)sin(5x) ?

The integral from 0 to pi of cos(6x)sin(5x) is -10/11

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integral from 0 to pi of cos(6x)sin(5x)

\int_{0}^{π} cos (6 x) sin (5 x) d x

- \frac{10}{11}

Decimal Notation

- 0.90909 \dots

Solution steps

\int_{0}^{π} cos (6 x) sin (5 x) d x

Use the following identity:

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

= \int_{0}^{π} \frac{sin ( 5 x + 6 x ) + sin ( 5 x - 6 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 (5 x + 6 x) + sin (5 x - 6 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 (5 x + 6 x) d x + \int_{0}^{π} sin (5 x - 6 x) d x)

\int_{0}^{π} sin (5 x + 6 x) d x = \frac{2}{11}

\int_{0}^{π} sin (5 x - 6 x) d x = - 2

= \frac{1}{2} (\frac{2}{11} - 2)

Simplify

\frac{1}{2} (\frac{2}{11} - 2) : - \frac{10}{11}

= - \frac{10}{11}

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What is the integral from 0 to pi of cos(6x)sin(5x) ?
The integral from 0 to pi of cos(6x)sin(5x) is -10/11