What is the integral from 0 to 1 of xcos(nx) ?

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

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integral from 0 to 1 of xcos(nx)

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

\frac{n sin ( n ) + cos ( n ) - 1}{n ^{2}}

Solution steps

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

Apply u-substitution

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

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

n sin (n) + cos (n) - 1

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

Simplify

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

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