Integraali $$$\frac{\sin^{2}{\left(z \right)}}{- \frac{\pi}{6} + z}$$$:stä muuttujan $$$\pi$$$ suhteen

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

Sovella vakiokertoimen sääntöä $$$\int c f{\left(\pi \right)}\, d\pi = c \int f{\left(\pi \right)}\, d\pi$$$ käyttäen $$$c=\sin^{2}{\left(z \right)}$$$ ja $$$f{\left(\pi \right)} = \frac{1}{- \frac{\pi}{6} + z}$$$:

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

Olkoon $$$u=- \frac{\pi}{6} + z$$$.

Tällöin $$$du=\left(- \frac{\pi}{6} + z\right)^{\prime }d\pi = - \frac{d\pi}{6}$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$d\pi = - 6 du$$$.

Näin ollen,

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

Sovella vakiokertoimen sääntöä $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ käyttäen $$$c=-6$$$ ja $$$f{\left(u \right)} = \frac{1}{u}$$$:

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

Funktion $$$\frac{1}{u}$$$ integraali on $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Muista, että $$$u=- \frac{\pi}{6} + z$$$:

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

Näin ollen,

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

Sievennä:

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

Lisää integrointivakio:

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

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