What is the integral from 0 to pi/2 of x*sin(nx) ?

The integral from 0 to pi/2 of x*sin(nx) is (2sin((pin)/2)-pincos((pin)/2))/(2n^2)

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integral from 0 to pi/2 of x*sin(nx)

\int_{0}^{\frac{π}{2}} x \cdot sin (n x) d x

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

Solution steps

\int_{0}^{\frac{π}{2}} x sin (n x) d x

Apply Integration By Parts

= [\frac{1}{n} (- x cos (n x) - n \cdot \int - \frac{1}{n} cos (n x) d x)]_{0}^{\frac{π}{2}}

\int - \frac{1}{n} cos (n x) d x = - \frac{1}{n ^{2}} sin (n x)

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

Simplify

[\frac{1}{n} (- x cos (n x) - n (- \frac{1}{n ^{2}} sin (n x)))]_{0}^{\frac{π}{2}} : [\frac{1}{n} (- x cos (n x) + \frac{1}{n} sin (n x))]_{0}^{\frac{π}{2}}

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

Compute the boundaries:

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

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

Simplify

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

What is the integral from 0 to pi/2 of x*sin(nx) ?
The integral from 0 to pi/2 of x*sin(nx) is (2sin((pin)/2)-pincos((pin)/2))/(2n^2)