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

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

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

\int_{0}^{π} cos (2 x) sin (n x) d x

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

Solution steps

\int_{0}^{π} cos (2 x) sin (n x) d x

Use the following identity:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

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

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{π} sin (n x + 2 x) + sin (n x - 2 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{0}^{π} sin (n x + 2 x) d x + \int_{0}^{π} sin (n x - 2 x) d x)

\int_{0}^{π} sin (n x + 2 x) d x = \frac{- ( - 1 ) ^{n} + 1}{n + 2}

\int_{0}^{π} sin (n x - 2 x) d x = \frac{- ( - 1 ) ^{n} + 1}{n - 2}

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

Simplify

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

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

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