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

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

4 x^{3} - 13 x^{2} + 4

9 (sin (x) + x cos (x))

f (t) = (3 t - 2)^{5}

y = x^{3}, 0 \leq x \leq 5

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