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

The integral from 0 to pi of sin(x)cos(kx) is -((-1)^{-k}+1)/((k+1)(k-1))

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integral from 0 to pi of sin(x)cos(kx)

\int_{0}^{π} sin (x) cos (k x) d x

- \frac{( - 1 ) ^{- k} + 1}{( k + 1 ) ( k - 1 )}

Solution steps

\int_{0}^{π} sin (x) cos (k x) d x

Use the following identity:

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

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

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

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

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

Simplify

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

= - \frac{( - 1 ) ^{- k} + 1}{( k + 1 ) ( k - 1 )}

\int_{0}^{l o g_{2} (6)} 2^{x} d x

\int_{- 1}^{2} (x^{2} + 1)^{2} d x

\int_{0}^{2 π} - 16 sin^{2} (θ) d θ

\int_{0}^{1} 2 y e^{- y^{2}} d y

\int_{0}^{2} x + y d y

What is the integral from 0 to pi of sin(x)cos(kx) ?
The integral from 0 to pi of sin(x)cos(kx) is -((-1)^{-k}+1)/((k+1)(k-1))