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

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

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integral from 0 to 2pi of sin(nθ)

\int_{0}^{2 π} sin (n θ) d θ

0

Solution steps

\int_{0}^{2 π} sin (n θ) d θ

Apply u-substitution

: \int_{0}^{n \cdot 2 π} \frac{sin ( u )}{n} d u

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{n} [- cos (u)]_{0}^{n \cdot 2 π}

Compute the boundaries:

0

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

Simplify

= 0

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

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