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

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

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

\int_{0}^{2 π} sin (2 x) d x

0

Solution steps

\int_{0}^{2 π} sin (2 x) d x

Apply u-substitution

: \int_{0}^{4 π} sin (u) \frac{1}{2} d u

= \int_{0}^{4 π} sin (u) \frac{1}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{4 π} sin (u) d u

Use the common integral:

\int sin (u) d u = - cos (u)

= \frac{1}{2} [- cos (u)]_{0}^{4 π}

Compute the boundaries:

0

= \frac{1}{2} \cdot 0

Simplify

= 0

\int \frac{e ^{2 x + 1}}{2} d x

ln (x^{2} + y^{2})

\frac{d}{d x} (e^{7 - x})

\frac{d ^{2}}{d x ^{2}} (9 t^{2} - 6 t + 3)

\frac{d}{d x} (ln (tanh (x)))

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