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

The integral from-pi to pi/2 of xcos(nx) is (pinsin((pin)/2)+2cos((pin)/2)-2(-1)^n)/(2n^2)

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integral from-pi to pi/2 of xcos(nx)

\int_{- π}^{\frac{π}{2}} x cos (n x) d x

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

Solution steps

\int_{- π}^{\frac{π}{2}} x cos (n x) d x

Apply u-substitution

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

= \int_{- πn}^{\frac{πn}{2}} \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_{- πn}^{\frac{πn}{2}} u cos (u) d u

Apply Integration By Parts

: [u sin (u) - \int sin (u) d u]_{- πn}^{\frac{πn}{2}}

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

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

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

Simplify

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

Compute the boundaries:

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

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

Simplify

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

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