Integrale di $$$\frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)} \cos{\left(4 t \right)}}{20}$$$

Trova $$$\int \frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)} \cos{\left(4 t \right)}}{20}\, dt$$$.

Riscrivi $$$\sin\left(2 t \right)\cos\left(4 t \right)$$$ 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 $$$\alpha=2 t$$$ e $$$\beta=4 t$$$:

$${\color{red}{\int{\frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)} \cos{\left(4 t \right)}}{20} d t}}} = {\color{red}{\int{\frac{\pi \left(- \frac{\sin{\left(2 t \right)}}{2} + \frac{\sin{\left(6 t \right)}}{2}\right) \sin{\left(4 t \right)}}{20} d t}}}$$

Sviluppa l'espressione:

$${\color{red}{\int{\frac{\pi \left(- \frac{\sin{\left(2 t \right)}}{2} + \frac{\sin{\left(6 t \right)}}{2}\right) \sin{\left(4 t \right)}}{20} d t}}} = {\color{red}{\int{\left(- \frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)}}{40} + \frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{40}\right)d t}}}$$

Applica la regola del fattore costante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ con $$$c=\frac{1}{2}$$$ e $$$f{\left(t \right)} = - \frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)}}{20} + \frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20}$$$:

$${\color{red}{\int{\left(- \frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)}}{40} + \frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{40}\right)d t}}} = {\color{red}{\left(\frac{\int{\left(- \frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)}}{20} + \frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20}\right)d t}}{2}\right)}}$$

Integra termine per termine:

$$\frac{{\color{red}{\int{\left(- \frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)}}{20} + \frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20}\right)d t}}}}{2} = \frac{{\color{red}{\left(- \int{\frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)}}{20} d t} + \int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}\right)}}}{2}$$

Riscrivi $$$\sin\left(2 t \right)\sin\left(4 t \right)$$$ utilizzando la formula $$$\sin\left(\alpha \right)\sin\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)-\frac{1}{2} \cos\left(\alpha+\beta \right)$$$ con $$$\alpha=2 t$$$ e $$$\beta=4 t$$$:

$$\frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{{\color{red}{\int{\frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)}}{20} d t}}}}{2} = \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{{\color{red}{\int{\frac{\pi \left(\frac{\cos{\left(2 t \right)}}{2} - \frac{\cos{\left(6 t \right)}}{2}\right)}{20} d t}}}}{2}$$

Sviluppa l'espressione:

$$\frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{{\color{red}{\int{\frac{\pi \left(\frac{\cos{\left(2 t \right)}}{2} - \frac{\cos{\left(6 t \right)}}{2}\right)}{20} d t}}}}{2} = \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{{\color{red}{\int{\left(\frac{\pi \cos{\left(2 t \right)}}{40} - \frac{\pi \cos{\left(6 t \right)}}{40}\right)d t}}}}{2}$$

Applica la regola del fattore costante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ con $$$c=\frac{1}{2}$$$ e $$$f{\left(t \right)} = \frac{\pi \cos{\left(2 t \right)}}{20} - \frac{\pi \cos{\left(6 t \right)}}{20}$$$:

$$\frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{{\color{red}{\int{\left(\frac{\pi \cos{\left(2 t \right)}}{40} - \frac{\pi \cos{\left(6 t \right)}}{40}\right)d t}}}}{2} = \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{{\color{red}{\left(\frac{\int{\left(\frac{\pi \cos{\left(2 t \right)}}{20} - \frac{\pi \cos{\left(6 t \right)}}{20}\right)d t}}{2}\right)}}}{2}$$

Integra termine per termine:

$$\frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{{\color{red}{\int{\left(\frac{\pi \cos{\left(2 t \right)}}{20} - \frac{\pi \cos{\left(6 t \right)}}{20}\right)d t}}}}{4} = \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{{\color{red}{\left(\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t} - \int{\frac{\pi \cos{\left(6 t \right)}}{20} d t}\right)}}}{4}$$

Applica la regola del fattore costante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ con $$$c=\frac{\pi}{20}$$$ e $$$f{\left(t \right)} = \cos{\left(6 t \right)}$$$:

$$- \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} + \frac{{\color{red}{\int{\frac{\pi \cos{\left(6 t \right)}}{20} d t}}}}{4} = - \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} + \frac{{\color{red}{\left(\frac{\pi \int{\cos{\left(6 t \right)} d t}}{20}\right)}}}{4}$$

Sia $$$u=6 t$$$.

Quindi $$$du=\left(6 t\right)^{\prime }dt = 6 dt$$$ (i passaggi si possono vedere »), e si ha che $$$dt = \frac{du}{6}$$$.

Quindi,

$$- \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} + \frac{\pi {\color{red}{\int{\cos{\left(6 t \right)} d t}}}}{80} = - \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} + \frac{\pi {\color{red}{\int{\frac{\cos{\left(u \right)}}{6} d u}}}}{80}$$

Applica la regola del fattore costante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{6}$$$ e $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$- \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} + \frac{\pi {\color{red}{\int{\frac{\cos{\left(u \right)}}{6} d u}}}}{80} = - \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} + \frac{\pi {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{6}\right)}}}{80}$$

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

$$- \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} + \frac{\pi {\color{red}{\int{\cos{\left(u \right)} d u}}}}{480} = - \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} + \frac{\pi {\color{red}{\sin{\left(u \right)}}}}{480}$$

Ricordiamo che $$$u=6 t$$$:

$$- \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} + \frac{\pi \sin{\left({\color{red}{u}} \right)}}{480} = - \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} + \frac{\pi \sin{\left({\color{red}{\left(6 t\right)}} \right)}}{480}$$

Applica la regola del fattore costante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ con $$$c=\frac{\pi}{20}$$$ e $$$f{\left(t \right)} = \cos{\left(2 t \right)}$$$:

$$\frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{{\color{red}{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}}}{4} = \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{{\color{red}{\left(\frac{\pi \int{\cos{\left(2 t \right)} d t}}{20}\right)}}}{4}$$

Sia $$$u=2 t$$$.

Quindi $$$du=\left(2 t\right)^{\prime }dt = 2 dt$$$ (i passaggi si possono vedere »), e si ha che $$$dt = \frac{du}{2}$$$.

Pertanto,

$$\frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{\pi {\color{red}{\int{\cos{\left(2 t \right)} d t}}}}{80} = \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{\pi {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{80}$$

Applica la regola del fattore costante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{2}$$$ e $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$\frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{\pi {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{80} = \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{\pi {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{80}$$

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

$$\frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{\pi {\color{red}{\int{\cos{\left(u \right)} d u}}}}{160} = \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{\pi {\color{red}{\sin{\left(u \right)}}}}{160}$$

Ricordiamo che $$$u=2 t$$$:

$$\frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{\pi \sin{\left({\color{red}{u}} \right)}}{160} = \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{\pi \sin{\left({\color{red}{\left(2 t\right)}} \right)}}{160}$$

Riscrivi $$$\sin\left(4 t \right)\sin\left(6 t \right)$$$ utilizzando la formula $$$\sin\left(\alpha \right)\sin\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)-\frac{1}{2} \cos\left(\alpha+\beta \right)$$$ con $$$\alpha=4 t$$$ e $$$\beta=6 t$$$:

$$- \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{{\color{red}{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}}}{2} = - \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{{\color{red}{\int{\frac{\pi \left(\frac{\cos{\left(2 t \right)}}{2} - \frac{\cos{\left(10 t \right)}}{2}\right)}{20} d t}}}}{2}$$

Sviluppa l'espressione:

$$- \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{{\color{red}{\int{\frac{\pi \left(\frac{\cos{\left(2 t \right)}}{2} - \frac{\cos{\left(10 t \right)}}{2}\right)}{20} d t}}}}{2} = - \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{{\color{red}{\int{\left(\frac{\pi \cos{\left(2 t \right)}}{40} - \frac{\pi \cos{\left(10 t \right)}}{40}\right)d t}}}}{2}$$

Applica la regola del fattore costante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ con $$$c=\frac{1}{2}$$$ e $$$f{\left(t \right)} = \frac{\pi \cos{\left(2 t \right)}}{20} - \frac{\pi \cos{\left(10 t \right)}}{20}$$$:

$$- \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{{\color{red}{\int{\left(\frac{\pi \cos{\left(2 t \right)}}{40} - \frac{\pi \cos{\left(10 t \right)}}{40}\right)d t}}}}{2} = - \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{{\color{red}{\left(\frac{\int{\left(\frac{\pi \cos{\left(2 t \right)}}{20} - \frac{\pi \cos{\left(10 t \right)}}{20}\right)d t}}{2}\right)}}}{2}$$

Integra termine per termine:

$$- \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{{\color{red}{\int{\left(\frac{\pi \cos{\left(2 t \right)}}{20} - \frac{\pi \cos{\left(10 t \right)}}{20}\right)d t}}}}{4} = - \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{{\color{red}{\left(\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t} - \int{\frac{\pi \cos{\left(10 t \right)}}{20} d t}\right)}}}{4}$$

Applica la regola del fattore costante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ con $$$c=\frac{\pi}{20}$$$ e $$$f{\left(t \right)} = \cos{\left(10 t \right)}$$$:

$$- \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} - \frac{{\color{red}{\int{\frac{\pi \cos{\left(10 t \right)}}{20} d t}}}}{4} = - \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} - \frac{{\color{red}{\left(\frac{\pi \int{\cos{\left(10 t \right)} d t}}{20}\right)}}}{4}$$

Sia $$$u=10 t$$$.

Quindi $$$du=\left(10 t\right)^{\prime }dt = 10 dt$$$ (i passaggi si possono vedere »), e si ha che $$$dt = \frac{du}{10}$$$.

Quindi,

$$- \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} - \frac{\pi {\color{red}{\int{\cos{\left(10 t \right)} d t}}}}{80} = - \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} - \frac{\pi {\color{red}{\int{\frac{\cos{\left(u \right)}}{10} d u}}}}{80}$$

Applica la regola del fattore costante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{10}$$$ e $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$- \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} - \frac{\pi {\color{red}{\int{\frac{\cos{\left(u \right)}}{10} d u}}}}{80} = - \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} - \frac{\pi {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{10}\right)}}}{80}$$

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

$$- \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} - \frac{\pi {\color{red}{\int{\cos{\left(u \right)} d u}}}}{800} = - \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} - \frac{\pi {\color{red}{\sin{\left(u \right)}}}}{800}$$

Ricordiamo che $$$u=10 t$$$:

$$- \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} - \frac{\pi \sin{\left({\color{red}{u}} \right)}}{800} = - \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} - \frac{\pi \sin{\left({\color{red}{\left(10 t\right)}} \right)}}{800}$$

Applica la regola del fattore costante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ con $$$c=\frac{\pi}{20}$$$ e $$$f{\left(t \right)} = \cos{\left(2 t \right)}$$$:

$$- \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} - \frac{\pi \sin{\left(10 t \right)}}{800} + \frac{{\color{red}{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}}}{4} = - \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} - \frac{\pi \sin{\left(10 t \right)}}{800} + \frac{{\color{red}{\left(\frac{\pi \int{\cos{\left(2 t \right)} d t}}{20}\right)}}}{4}$$

L'integrale $$$\int{\cos{\left(2 t \right)} d t}$$$ è già stato calcolato:

$$\int{\cos{\left(2 t \right)} d t} = \frac{\sin{\left(2 t \right)}}{2}$$

Pertanto,

$$- \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} - \frac{\pi \sin{\left(10 t \right)}}{800} + \frac{\pi {\color{red}{\int{\cos{\left(2 t \right)} d t}}}}{80} = - \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} - \frac{\pi \sin{\left(10 t \right)}}{800} + \frac{\pi {\color{red}{\left(\frac{\sin{\left(2 t \right)}}{2}\right)}}}{80}$$

Pertanto,

$$\int{\frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)} \cos{\left(4 t \right)}}{20} d t} = \frac{\pi \sin{\left(6 t \right)}}{480} - \frac{\pi \sin{\left(10 t \right)}}{800}$$

Semplifica:

$$\int{\frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)} \cos{\left(4 t \right)}}{20} d t} = \frac{\pi \left(5 \sin{\left(6 t \right)} - 3 \sin{\left(10 t \right)}\right)}{2400}$$

Aggiungi la costante di integrazione:

$$\int{\frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)} \cos{\left(4 t \right)}}{20} d t} = \frac{\pi \left(5 \sin{\left(6 t \right)} - 3 \sin{\left(10 t \right)}\right)}{2400}+C$$

$$$\int \frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)} \cos{\left(4 t \right)}}{20}\, dt = \frac{\pi \left(5 \sin{\left(6 t \right)} - 3 \sin{\left(10 t \right)}\right)}{2400} + C$$$A