What is the integral from 0 to pi of sin(6x)cos(3x) ?

The integral from 0 to pi of sin(6x)cos(3x) is 4/9

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integral from 0 to pi of sin(6x)cos(3x)

\int_{0}^{π} sin (6 x) cos (3 x) d x

\frac{4}{9}

Decimal Notation

0.44444 \dots

Solution steps

\int_{0}^{π} sin (6 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 ( 6 x + 3 x ) + sin ( 6 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 (6 x + 3 x) + sin (6 x - 3 x) d x

Simplify

= \frac{1}{2} \cdot \int_{0}^{π} sin (9 x) + sin (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 (9 x) d x + \int_{0}^{π} sin (3 x) d x)

\int_{0}^{π} sin (9 x) d x = \frac{2}{9}

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

= \frac{1}{2} (\frac{2}{9} + \frac{2}{3})

Simplify

\frac{1}{2} (\frac{2}{9} + \frac{2}{3}) : \frac{4}{9}

= \frac{4}{9}

\int - 2 e^{x} (e^{x} - 1)^{3} d x

9 arctan (x - 1 + x^{2})

(19, - 16), (- 7, - 15)

x \to 2 lim (x^{3})

7 x e^{x}

What is the integral from 0 to pi of sin(6x)cos(3x) ?
The integral from 0 to pi of sin(6x)cos(3x) is 4/9