integral from 0 to pi/2 of sin(x)cos(3x)

Lösung

\int_{0}^{\frac{π}{2}} sin (x) cos (3 x) d x

- \frac{1}{2}

Dezimale

- 0.5

Schritte zur Lösung

\int_{0}^{\frac{π}{2}} sin (x) cos (3 x) d x

Verwende die folgenden Identitäten:

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

= \int_{0}^{\frac{π}{2}} \frac{sin ( x + 3 x ) + sin ( x - 3 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_{0}^{\frac{π}{2}} sin (x + 3 x) + sin (x - 3 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_{0}^{\frac{π}{2}} sin (x + 3 x) d x + \int_{0}^{\frac{π}{2}} sin (x - 3 x) d x)

\int_{0}^{\frac{π}{2}} sin (x + 3 x) d x = 0

\int_{0}^{\frac{π}{2}} sin (x - 3 x) d x = - 1

= \frac{1}{2} (0 - 1)

Vereinfache

= - \frac{1}{2}

\int_{\frac{1}{2}}^{\frac{1}{3}} cos (π) x d x

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

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\int_{x}^{0} x^{2} + 4 d x