What is the integral from-pi to 0 of-1cos(nx) ?

The integral from-pi to 0 of-1cos(nx) is 0

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integral from-pi to 0 of-1cos(nx)

\int_{- π}^{0} - 1 cos (n x) d x

0

Solution steps

\int_{- π}^{0} - 1 \cdot cos (n x) d x

Take the constant out:

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

= - \int_{- π}^{0} cos (n x) d x

Apply u-substitution

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

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

Use the common integral:

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

= - \frac{1}{n} [sin (u)]_{- πn}^{0}

Compute the boundaries:

0

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

Simplify

= 0

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What is the integral from-pi to 0 of-1cos(nx) ?
The integral from-pi to 0 of-1cos(nx) is 0