What is the integral from 0 to pi of zcos(nz) ?

The integral from 0 to pi of zcos(nz) is ((-1)^n-1)/(n^2)

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integral from 0 to pi of zcos(nz)

\int_{0}^{π} z cos (n z) d z

\frac{( - 1 ) ^{n} - 1}{n ^{2}}

Solution steps

\int_{0}^{π} z cos (n z) d z

Apply u-substitution

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

Take the constant out:

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

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

Apply Integration By Parts

= \frac{1}{n ^{2}} [u sin (u) - \int sin (u) d u]_{0}^{πn}

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

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

Simplify

= \frac{1}{n ^{2}} [u sin (u) + cos (u)]_{0}^{πn}

Compute the boundaries:

(- 1)^{n} - 1

= \frac{1}{n ^{2}} ((- 1)^{n} - 1)

Simplify

= \frac{( - 1 ) ^{n} - 1}{n ^{2}}

What is the integral from 0 to pi of zcos(nz) ?
The integral from 0 to pi of zcos(nz) is ((-1)^n-1)/(n^2)