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

Lösung

\int_{- π}^{0} sin (x) cos (n x) d x

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

Schritte zur Lösung

\int_{- π}^{0} sin (x) cos (n x) d x

Verwende die folgenden Identitäten:

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

Entferne die Konstante:

\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

Wende die Summenregel an:

\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})

Vereinfache

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

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

\int_{0}^{1} 2 x^{3} 1 - x^{2} d x

\int_{3}^{5} \frac{1}{x ^{2} - 4} d x