What is the integral from 0 to 1 of cos(npit) ?

The integral from 0 to 1 of cos(npit) is 0

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integral from 0 to 1 of cos(npit)

\int_{0}^{1} cos (nπ t) d t

0

Solution steps

\int_{0}^{1} cos (nπ t) d t

Apply u-substitution

: \int_{0}^{πn} \frac{cos ( u )}{πn} d u

= \int_{0}^{πn} \frac{cos ( u )}{πn} d u

Take the constant out:

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

= \frac{1}{πn} \cdot \int_{0}^{πn} cos (u) d u

Use the common integral:

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

= \frac{1}{πn} [sin (u)]_{0}^{πn}

Compute the boundaries:

0

= \frac{1}{πn} \cdot 0

Simplify

= 0

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What is the integral from 0 to 1 of cos(npit) ?
The integral from 0 to 1 of cos(npit) is 0