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

The integral from 0 to 2pi of sin(2x)sin(x) is 0

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

\int_{0}^{2 π} sin (2 x) sin (x) d x

0

Solution steps

\int_{0}^{2 π} sin (2 x) sin (x) d x

Rewrite using trig identities

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

\int_{0}^{2 π} cos (3 x) d x = 0

\int_{0}^{2 π} cos (x) d x = 0

= \frac{1}{2} (- 0 + 0)

Simplify

= 0

n = 0 \sum \infty (\frac{- 1}{2})^{n}

\int x^{5} - 5 x^{- 2} + 2 d x

f (x) = 4 x (x^{2} - 4 x + 5)^{9}, at x = 2

\frac{d}{d x} (9 e^{- 4 x})

y^{3} \frac{d y}{d x} + x^{3} = 0

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