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

The integral from 0 to pi of sin(x)cos(4x) is -2/15

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

\int_{0}^{π} sin (x) cos (4 x) d x

- \frac{2}{15}

Decimal Notation

- 0.13333 \dots

Solution steps

\int_{0}^{π} sin (x) cos (4 x) d x

Use the following identity:

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

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

\int_{0}^{π} sin (x + 4 x) d x = \frac{2}{5}

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

= \frac{1}{2} (\frac{2}{5} - \frac{2}{3})

Simplify

\frac{1}{2} (\frac{2}{5} - \frac{2}{3}) : - \frac{2}{15}

= - \frac{2}{15}

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What is the integral from 0 to pi of sin(x)cos(4x) ?
The integral from 0 to pi of sin(x)cos(4x) is -2/15