$$$\pi$$$에 대한 $$$\frac{\sin^{2}{\left(z \right)}}{- \frac{\pi}{6} + z}$$$의 적분

$$$\int \frac{\sin^{2}{\left(z \right)}}{- \frac{\pi}{6} + z}\, d\pi$$$을(를) 구하시오.

상수배 법칙 $$$\int c f{\left(\pi \right)}\, d\pi = c \int f{\left(\pi \right)}\, d\pi$$$을 $$$c=\sin^{2}{\left(z \right)}$$$와 $$$f{\left(\pi \right)} = \frac{1}{- \frac{\pi}{6} + z}$$$에 적용하세요:

$${\color{red}{\int{\frac{\sin^{2}{\left(z \right)}}{- \frac{\pi}{6} + z} d \pi}}} = {\color{red}{\sin^{2}{\left(z \right)} \int{\frac{1}{- \frac{\pi}{6} + z} d \pi}}}$$

$$$u=- \frac{\pi}{6} + z$$$라 하자.

그러면 $$$du=\left(- \frac{\pi}{6} + z\right)^{\prime }d\pi = - \frac{d\pi}{6}$$$ (단계는 »에서 볼 수 있습니다), 그리고 $$$d\pi = - 6 du$$$임을 얻습니다.

따라서,

$$\sin^{2}{\left(z \right)} {\color{red}{\int{\frac{1}{- \frac{\pi}{6} + z} d \pi}}} = \sin^{2}{\left(z \right)} {\color{red}{\int{\left(- \frac{6}{u}\right)d u}}}$$

상수배 법칙 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$을 $$$c=-6$$$와 $$$f{\left(u \right)} = \frac{1}{u}$$$에 적용하세요:

$$\sin^{2}{\left(z \right)} {\color{red}{\int{\left(- \frac{6}{u}\right)d u}}} = \sin^{2}{\left(z \right)} {\color{red}{\left(- 6 \int{\frac{1}{u} d u}\right)}}$$

$$$\frac{1}{u}$$$의 적분은 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$- 6 \sin^{2}{\left(z \right)} {\color{red}{\int{\frac{1}{u} d u}}} = - 6 \sin^{2}{\left(z \right)} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

다음 $$$u=- \frac{\pi}{6} + z$$$을 기억하라:

$$- 6 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} \sin^{2}{\left(z \right)} = - 6 \ln{\left(\left|{{\color{red}{\left(- \frac{\pi}{6} + z\right)}}}\right| \right)} \sin^{2}{\left(z \right)}$$

따라서,

$$\int{\frac{\sin^{2}{\left(z \right)}}{- \frac{\pi}{6} + z} d \pi} = - 6 \ln{\left(\left|{\frac{\pi}{6} - z}\right| \right)} \sin^{2}{\left(z \right)}$$

간단히 하시오:

$$\int{\frac{\sin^{2}{\left(z \right)}}{- \frac{\pi}{6} + z} d \pi} = 6 \left(- \ln{\left(\left|{\pi - 6 z}\right| \right)} + \ln{\left(6 \right)}\right) \sin^{2}{\left(z \right)}$$

적분 상수를 추가하세요:

$$\int{\frac{\sin^{2}{\left(z \right)}}{- \frac{\pi}{6} + z} d \pi} = 6 \left(- \ln{\left(\left|{\pi - 6 z}\right| \right)} + \ln{\left(6 \right)}\right) \sin^{2}{\left(z \right)}+C$$

$$$\int \frac{\sin^{2}{\left(z \right)}}{- \frac{\pi}{6} + z}\, d\pi = 6 \left(- \ln\left(\left|{\pi - 6 z}\right|\right) + \ln\left(6\right)\right) \sin^{2}{\left(z \right)} + C$$$A