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

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

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integral from 0 to 2pi of (sin(θ))

\int_{0}^{2 π} (sin (θ)) d θ

0

Solution steps

\int_{0}^{2 π} sin (θ) d θ

Use the common integral:

\int sin (θ) d θ = - cos (θ)

= [- cos (θ)]_{0}^{2 π}

Compute the boundaries:

0

= 0

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\int_{- \infty}^{0} \frac{x}{x ^{4} + 1} d x

\int_{- 1}^{0} \frac{1}{1 - x} d x

\int_{1}^{6} 4 x^{3} - 4 x^{4} - 3 d x

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