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

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

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integral from 0 to pi of 2xcos(nx)

\int_{0}^{π} 2 x cos (n x) d x

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

Solution steps

\int_{0}^{π} 2 x cos (n x) d x

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \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

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

Apply Integration By Parts

: [u sin (u) - \int sin (u) d u]_{0}^{πn}

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

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

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

Simplify

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

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

Compute the boundaries:

(- 1)^{n} - 1

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

Simplify

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

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