Integralen av $$$t \sin{\left(t \right)} \cos{\left(t \right)}$$$

Bestäm $$$\int t \sin{\left(t \right)} \cos{\left(t \right)}\, dt$$$.

För integralen $$$\int{t \sin{\left(t \right)} \cos{\left(t \right)} d t}$$$, använd partiell integration $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Låt $$$\operatorname{u}=t$$$ och $$$\operatorname{dv}=\sin{\left(t \right)} \cos{\left(t \right)} dt$$$.

Då gäller $$$\operatorname{du}=\left(t\right)^{\prime }dt=1 dt$$$ (stegen kan ses ») och $$$\operatorname{v}=\int{\sin{\left(t \right)} \cos{\left(t \right)} d t}=\frac{\sin^{2}{\left(t \right)}}{2}$$$ (stegen kan ses »).

Alltså,

$${\color{red}{\int{t \sin{\left(t \right)} \cos{\left(t \right)} d t}}}={\color{red}{\left(t \cdot \frac{\sin^{2}{\left(t \right)}}{2}-\int{\frac{\sin^{2}{\left(t \right)}}{2} \cdot 1 d t}\right)}}={\color{red}{\left(\frac{t \sin^{2}{\left(t \right)}}{2} - \int{\frac{\sin^{2}{\left(t \right)}}{2} d t}\right)}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ med $$$c=\frac{1}{2}$$$ och $$$f{\left(t \right)} = \sin^{2}{\left(t \right)}$$$:

$$\frac{t \sin^{2}{\left(t \right)}}{2} - {\color{red}{\int{\frac{\sin^{2}{\left(t \right)}}{2} d t}}} = \frac{t \sin^{2}{\left(t \right)}}{2} - {\color{red}{\left(\frac{\int{\sin^{2}{\left(t \right)} d t}}{2}\right)}}$$

Använd potensreduceringsformeln $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$ med $$$\alpha=t$$$:

$$\frac{t \sin^{2}{\left(t \right)}}{2} - \frac{{\color{red}{\int{\sin^{2}{\left(t \right)} d t}}}}{2} = \frac{t \sin^{2}{\left(t \right)}}{2} - \frac{{\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 t \right)}}{2}\right)d t}}}}{2}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ med $$$c=\frac{1}{2}$$$ och $$$f{\left(t \right)} = 1 - \cos{\left(2 t \right)}$$$:

$$\frac{t \sin^{2}{\left(t \right)}}{2} - \frac{{\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 t \right)}}{2}\right)d t}}}}{2} = \frac{t \sin^{2}{\left(t \right)}}{2} - \frac{{\color{red}{\left(\frac{\int{\left(1 - \cos{\left(2 t \right)}\right)d t}}{2}\right)}}}{2}$$

Integrera termvis:

$$\frac{t \sin^{2}{\left(t \right)}}{2} - \frac{{\color{red}{\int{\left(1 - \cos{\left(2 t \right)}\right)d t}}}}{4} = \frac{t \sin^{2}{\left(t \right)}}{2} - \frac{{\color{red}{\left(\int{1 d t} - \int{\cos{\left(2 t \right)} d t}\right)}}}{4}$$

Tillämpa konstantregeln $$$\int c\, dt = c t$$$ med $$$c=1$$$:

$$\frac{t \sin^{2}{\left(t \right)}}{2} + \frac{\int{\cos{\left(2 t \right)} d t}}{4} - \frac{{\color{red}{\int{1 d t}}}}{4} = \frac{t \sin^{2}{\left(t \right)}}{2} + \frac{\int{\cos{\left(2 t \right)} d t}}{4} - \frac{{\color{red}{t}}}{4}$$

Låt $$$u=2 t$$$ vara.

Då $$$du=\left(2 t\right)^{\prime }dt = 2 dt$$$ (stegen kan ses »), och vi har att $$$dt = \frac{du}{2}$$$.

Alltså,

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{2}$$$ och $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

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

Integralen av cosinus är $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

$$\frac{t \sin^{2}{\left(t \right)}}{2} - \frac{t}{4} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{8} = \frac{t \sin^{2}{\left(t \right)}}{2} - \frac{t}{4} + \frac{{\color{red}{\sin{\left(u \right)}}}}{8}$$

Kom ihåg att $$$u=2 t$$$:

$$\frac{t \sin^{2}{\left(t \right)}}{2} - \frac{t}{4} + \frac{\sin{\left({\color{red}{u}} \right)}}{8} = \frac{t \sin^{2}{\left(t \right)}}{2} - \frac{t}{4} + \frac{\sin{\left({\color{red}{\left(2 t\right)}} \right)}}{8}$$

Alltså,

$$\int{t \sin{\left(t \right)} \cos{\left(t \right)} d t} = \frac{t \sin^{2}{\left(t \right)}}{2} - \frac{t}{4} + \frac{\sin{\left(2 t \right)}}{8}$$

Lägg till integrationskonstanten:

$$\int{t \sin{\left(t \right)} \cos{\left(t \right)} d t} = \frac{t \sin^{2}{\left(t \right)}}{2} - \frac{t}{4} + \frac{\sin{\left(2 t \right)}}{8}+C$$

$$$\int t \sin{\left(t \right)} \cos{\left(t \right)}\, dt = \left(\frac{t \sin^{2}{\left(t \right)}}{2} - \frac{t}{4} + \frac{\sin{\left(2 t \right)}}{8}\right) + C$$$A