Integralen av $$$2 \sin^{2}{\left(t \right)}$$$

Bestäm $$$\int 2 \sin^{2}{\left(t \right)}\, dt$$$.

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

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

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

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

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

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

Integrera termvis:

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

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

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

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,

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

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

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

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

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

Kom ihåg att $$$u=2 t$$$:

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

Alltså,

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

Lägg till integrationskonstanten:

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

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