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

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

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

\int_{0}^{π} sin (x) cos (2 x) d x

- \frac{2}{3}

Decimal Notation

- 0.66666 \dots

Solution steps

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

Simplify

sin (x + 2 x) + sin (x - 2 x) : sin (3 x) - sin (x)

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

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

\int_{0}^{π} sin (x) d x = 2

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

Simplify

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

= - \frac{2}{3}

\int_{- 1}^{2} (2 - x^{2} + x) d x

x \to 0 lim (x cos (\frac{11}{x}))

f (x) = - \frac{8}{x ^{3}}

\frac{d y}{d x} = x y

t \to 0 lim (e^{- 5 t})

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