$$$\left(1 - \sin{\left(\frac{t}{2} \right)}\right)^{2} \cos{\left(\frac{t}{2} \right)}$$$의 적분

$$$\int \left(1 - \sin{\left(\frac{t}{2} \right)}\right)^{2} \cos{\left(\frac{t}{2} \right)}\, dt$$$을(를) 구하시오.

$$$u=1 - \sin{\left(\frac{t}{2} \right)}$$$라 하자.

그러면 $$$du=\left(1 - \sin{\left(\frac{t}{2} \right)}\right)^{\prime }dt = - \frac{\cos{\left(\frac{t}{2} \right)}}{2} dt$$$ (단계는 »에서 볼 수 있습니다), 그리고 $$$\cos{\left(\frac{t}{2} \right)} dt = - 2 du$$$임을 얻습니다.

따라서,

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

상수배 법칙 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$을 $$$c=-2$$$와 $$$f{\left(u \right)} = u^{2}$$$에 적용하세요:

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

멱법칙($$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$)을 $$$n=2$$$에 적용합니다:

$$- 2 {\color{red}{\int{u^{2} d u}}}=- 2 {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=- 2 {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

다음 $$$u=1 - \sin{\left(\frac{t}{2} \right)}$$$을 기억하라:

$$- \frac{2 {\color{red}{u}}^{3}}{3} = - \frac{2 {\color{red}{\left(1 - \sin{\left(\frac{t}{2} \right)}\right)}}^{3}}{3}$$

따라서,

$$\int{\left(1 - \sin{\left(\frac{t}{2} \right)}\right)^{2} \cos{\left(\frac{t}{2} \right)} d t} = - \frac{2 \left(1 - \sin{\left(\frac{t}{2} \right)}\right)^{3}}{3}$$

간단히 하시오:

$$\int{\left(1 - \sin{\left(\frac{t}{2} \right)}\right)^{2} \cos{\left(\frac{t}{2} \right)} d t} = \frac{2 \left(\sin{\left(\frac{t}{2} \right)} - 1\right)^{3}}{3}$$

적분 상수를 추가하세요:

$$\int{\left(1 - \sin{\left(\frac{t}{2} \right)}\right)^{2} \cos{\left(\frac{t}{2} \right)} d t} = \frac{2 \left(\sin{\left(\frac{t}{2} \right)} - 1\right)^{3}}{3}+C$$

$$$\int \left(1 - \sin{\left(\frac{t}{2} \right)}\right)^{2} \cos{\left(\frac{t}{2} \right)}\, dt = \frac{2 \left(\sin{\left(\frac{t}{2} \right)} - 1\right)^{3}}{3} + C$$$A