What is the integral from 0 to 2pi of cos(nx) ?

The integral from 0 to 2pi of cos(nx) is (sin(2pin))/n

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integral from 0 to 2pi of cos(nx)

\int_{0}^{2 π} cos (n x) d x

\frac{sin ( 2 πn )}{n}

Solution steps

\int_{0}^{2 π} cos (n x) d x

Apply u-substitution

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

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

Use the common integral:

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

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

Compute the boundaries:

sin (2 πn)

= \frac{1}{n} sin (2 πn)

Simplify

= \frac{sin ( 2 πn )}{n}

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What is the integral from 0 to 2pi of cos(nx) ?
The integral from 0 to 2pi of cos(nx) is (sin(2pin))/n