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

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

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

\int_{- π}^{0} sin (x) cos (n x) d x

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

Solution steps

\int_{- π}^{0} sin (x) cos (n x) d x

Use the following identity:

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

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

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

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

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

Simplify

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

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

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

\int_{1}^{5} \frac{z ^{2} + 3}{z} d z

\int_{0}^{7} 14 x - x^{2} d x

\int_{0}^{6} \frac{x}{x ^{2} + 1} d x

\int_{1}^{8} t^{2} - 4 t - 12 d t

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