$$$\sin^{2}{\left(\alpha \right)}$$$'nin integrali

Bulun: $$$\int \sin^{2}{\left(\alpha \right)}\, d\alpha$$$.

Kuvvet indirgeme formülü $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$'i $$$\alpha=\alpha$$$ ile uygula:

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

Sabit katsayı kuralı $$$\int c f{\left(\alpha \right)}\, d\alpha = c \int f{\left(\alpha \right)}\, d\alpha$$$'i $$$c=\frac{1}{2}$$$ ve $$$f{\left(\alpha \right)} = 1 - \cos{\left(2 \alpha \right)}$$$ ile uygula:

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

Her terimin integralini alın:

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

$$$c=1$$$ kullanarak $$$\int c\, d\alpha = \alpha c$$$ sabit kuralını uygula:

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

$$$u=2 \alpha$$$ olsun.

Böylece $$$du=\left(2 \alpha\right)^{\prime }d\alpha = 2 d\alpha$$$ (adımlar » görülebilir) ve $$$d\alpha = \frac{du}{2}$$$ elde ederiz.

İntegral şu hale gelir

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

Sabit katsayı kuralı $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$'i $$$c=\frac{1}{2}$$$ ve $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ ile uygula:

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

Kosinüsün integrali $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

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

Hatırlayın ki $$$u=2 \alpha$$$:

$$\frac{\alpha}{2} - \frac{\sin{\left({\color{red}{u}} \right)}}{4} = \frac{\alpha}{2} - \frac{\sin{\left({\color{red}{\left(2 \alpha\right)}} \right)}}{4}$$

Dolayısıyla,

$$\int{\sin^{2}{\left(\alpha \right)} d \alpha} = \frac{\alpha}{2} - \frac{\sin{\left(2 \alpha \right)}}{4}$$

İntegrasyon sabitini ekleyin:

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

$$$\int \sin^{2}{\left(\alpha \right)}\, d\alpha = \left(\frac{\alpha}{2} - \frac{\sin{\left(2 \alpha \right)}}{4}\right) + C$$$A