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

The integral from 0 to pi of sin(x/2)cos(nx) is -(8n^2)/((2n+1)(2n-1))+2

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

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

- \frac{8 n ^{2}}{( 2 n + 1 ) ( 2 n - 1 )} + 2

Solution steps

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

- n \frac{2}{2 n + 1} - n \frac{2}{2 n - 1} + 2

= - n \frac{2}{2 n + 1} - n \frac{2}{2 n - 1} + 2

Simplify

= - \frac{8 n ^{2}}{( 2 n + 1 ) ( 2 n - 1 )} + 2

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What is the integral from 0 to pi of sin(x/2)cos(nx) ?
The integral from 0 to pi of sin(x/2)cos(nx) is -(8n^2)/((2n+1)(2n-1))+2