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

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

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

\int_{0}^{2 π} (sin (x)) d x

0

Solution steps

\int_{0}^{2 π} sin (x) d x

Use the common integral:

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

= [- cos (x)]_{0}^{2 π}

Compute the boundaries:

0

= 0

\int_{0}^{5} x d x

\int_{0}^{4} ∣ x - 1 ∣ d x

\int_{- 1}^{2} \frac{x}{x ^{2} + 4} d x

\int_{0}^{2} x^{3} 16 - x^{4} d x

\int_{0}^{3} \frac{x}{1 + 1 + x} d x

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