Integrale di $$$\sin{\left(2 \theta \right)} \cos{\left(\theta \right)}$$$

Trova $$$\int \sin{\left(2 \theta \right)} \cos{\left(\theta \right)}\, d\theta$$$.

Riscrivi l’integrando utilizzando la formula $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$ con $$$\alpha=2 \theta$$$ e $$$\beta=\theta$$$:

$${\color{red}{\int{\sin{\left(2 \theta \right)} \cos{\left(\theta \right)} d \theta}}} = {\color{red}{\int{\left(\frac{\sin{\left(\theta \right)}}{2} + \frac{\sin{\left(3 \theta \right)}}{2}\right)d \theta}}}$$

Applica la regola del fattore costante $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ con $$$c=\frac{1}{2}$$$ e $$$f{\left(\theta \right)} = \sin{\left(\theta \right)} + \sin{\left(3 \theta \right)}$$$:

$${\color{red}{\int{\left(\frac{\sin{\left(\theta \right)}}{2} + \frac{\sin{\left(3 \theta \right)}}{2}\right)d \theta}}} = {\color{red}{\left(\frac{\int{\left(\sin{\left(\theta \right)} + \sin{\left(3 \theta \right)}\right)d \theta}}{2}\right)}}$$

Integra termine per termine:

$$\frac{{\color{red}{\int{\left(\sin{\left(\theta \right)} + \sin{\left(3 \theta \right)}\right)d \theta}}}}{2} = \frac{{\color{red}{\left(\int{\sin{\left(\theta \right)} d \theta} + \int{\sin{\left(3 \theta \right)} d \theta}\right)}}}{2}$$

L'integrale del seno è $$$\int{\sin{\left(\theta \right)} d \theta} = - \cos{\left(\theta \right)}$$$:

$$\frac{\int{\sin{\left(3 \theta \right)} d \theta}}{2} + \frac{{\color{red}{\int{\sin{\left(\theta \right)} d \theta}}}}{2} = \frac{\int{\sin{\left(3 \theta \right)} d \theta}}{2} + \frac{{\color{red}{\left(- \cos{\left(\theta \right)}\right)}}}{2}$$

Sia $$$u=3 \theta$$$.

Quindi $$$du=\left(3 \theta\right)^{\prime }d\theta = 3 d\theta$$$ (i passaggi si possono vedere »), e si ha che $$$d\theta = \frac{du}{3}$$$.

Pertanto,

$$- \frac{\cos{\left(\theta \right)}}{2} + \frac{{\color{red}{\int{\sin{\left(3 \theta \right)} d \theta}}}}{2} = - \frac{\cos{\left(\theta \right)}}{2} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}}}{2}$$

Applica la regola del fattore costante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{3}$$$ e $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$- \frac{\cos{\left(\theta \right)}}{2} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}}}{2} = - \frac{\cos{\left(\theta \right)}}{2} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{3}\right)}}}{2}$$

L'integrale del seno è $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

$$- \frac{\cos{\left(\theta \right)}}{2} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{6} = - \frac{\cos{\left(\theta \right)}}{2} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{6}$$

Ricordiamo che $$$u=3 \theta$$$:

$$- \frac{\cos{\left(\theta \right)}}{2} - \frac{\cos{\left({\color{red}{u}} \right)}}{6} = - \frac{\cos{\left(\theta \right)}}{2} - \frac{\cos{\left({\color{red}{\left(3 \theta\right)}} \right)}}{6}$$

Pertanto,

$$\int{\sin{\left(2 \theta \right)} \cos{\left(\theta \right)} d \theta} = - \frac{\cos{\left(\theta \right)}}{2} - \frac{\cos{\left(3 \theta \right)}}{6}$$

Aggiungi la costante di integrazione:

$$\int{\sin{\left(2 \theta \right)} \cos{\left(\theta \right)} d \theta} = - \frac{\cos{\left(\theta \right)}}{2} - \frac{\cos{\left(3 \theta \right)}}{6}+C$$

$$$\int \sin{\left(2 \theta \right)} \cos{\left(\theta \right)}\, d\theta = \left(- \frac{\cos{\left(\theta \right)}}{2} - \frac{\cos{\left(3 \theta \right)}}{6}\right) + C$$$A