Funktion $$$2 \ln\left(\sin{\left(x \right)}\right) \sin^{6}{\left(x \right)} \cos{\left(x \right)}$$$ integraali
Aiheeseen liittyvä laskin: Määrättyjen ja epäoleellisten integraalien laskin
Syötteesi
Määritä $$$\int 2 \ln\left(\sin{\left(x \right)}\right) \sin^{6}{\left(x \right)} \cos{\left(x \right)}\, dx$$$.
Ratkaisu
Sovella vakiokertoimen sääntöä $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ käyttäen $$$c=2$$$ ja $$$f{\left(x \right)} = \ln{\left(\sin{\left(x \right)} \right)} \sin^{6}{\left(x \right)} \cos{\left(x \right)}$$$:
$${\color{red}{\int{2 \ln{\left(\sin{\left(x \right)} \right)} \sin^{6}{\left(x \right)} \cos{\left(x \right)} d x}}} = {\color{red}{\left(2 \int{\ln{\left(\sin{\left(x \right)} \right)} \sin^{6}{\left(x \right)} \cos{\left(x \right)} d x}\right)}}$$
Olkoon $$$u=\sin{\left(x \right)}$$$.
Tällöin $$$du=\left(\sin{\left(x \right)}\right)^{\prime }dx = \cos{\left(x \right)} dx$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$\cos{\left(x \right)} dx = du$$$.
Näin ollen,
$$2 {\color{red}{\int{\ln{\left(\sin{\left(x \right)} \right)} \sin^{6}{\left(x \right)} \cos{\left(x \right)} d x}}} = 2 {\color{red}{\int{u^{6} \ln{\left(u \right)} d u}}}$$
Integraalin $$$\int{u^{6} \ln{\left(u \right)} d u}$$$ kohdalla käytä osittaisintegrointia $$$\int \operatorname{\kappa} \operatorname{dv} = \operatorname{\kappa}\operatorname{v} - \int \operatorname{v} \operatorname{d\kappa}$$$.
Olkoon $$$\operatorname{\kappa}=\ln{\left(u \right)}$$$ ja $$$\operatorname{dv}=u^{6} du$$$.
Tällöin $$$\operatorname{d\kappa}=\left(\ln{\left(u \right)}\right)^{\prime }du=\frac{du}{u}$$$ (vaiheet ovat nähtävissä ») ja $$$\operatorname{v}=\int{u^{6} d u}=\frac{u^{7}}{7}$$$ (vaiheet ovat nähtävissä »).
Siis,
$$2 {\color{red}{\int{u^{6} \ln{\left(u \right)} d u}}}=2 {\color{red}{\left(\ln{\left(u \right)} \cdot \frac{u^{7}}{7}-\int{\frac{u^{7}}{7} \cdot \frac{1}{u} d u}\right)}}=2 {\color{red}{\left(\frac{u^{7} \ln{\left(u \right)}}{7} - \int{\frac{u^{6}}{7} d u}\right)}}$$
Sovella vakiokertoimen sääntöä $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ käyttäen $$$c=\frac{1}{7}$$$ ja $$$f{\left(u \right)} = u^{6}$$$:
$$\frac{2 u^{7} \ln{\left(u \right)}}{7} - 2 {\color{red}{\int{\frac{u^{6}}{7} d u}}} = \frac{2 u^{7} \ln{\left(u \right)}}{7} - 2 {\color{red}{\left(\frac{\int{u^{6} d u}}{7}\right)}}$$
Sovella potenssisääntöä $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ käyttäen $$$n=6$$$:
$$\frac{2 u^{7} \ln{\left(u \right)}}{7} - \frac{2 {\color{red}{\int{u^{6} d u}}}}{7}=\frac{2 u^{7} \ln{\left(u \right)}}{7} - \frac{2 {\color{red}{\frac{u^{1 + 6}}{1 + 6}}}}{7}=\frac{2 u^{7} \ln{\left(u \right)}}{7} - \frac{2 {\color{red}{\left(\frac{u^{7}}{7}\right)}}}{7}$$
Muista, että $$$u=\sin{\left(x \right)}$$$:
$$- \frac{2 {\color{red}{u}}^{7}}{49} + \frac{2 {\color{red}{u}}^{7} \ln{\left({\color{red}{u}} \right)}}{7} = - \frac{2 {\color{red}{\sin{\left(x \right)}}}^{7}}{49} + \frac{2 {\color{red}{\sin{\left(x \right)}}}^{7} \ln{\left({\color{red}{\sin{\left(x \right)}}} \right)}}{7}$$
Näin ollen,
$$\int{2 \ln{\left(\sin{\left(x \right)} \right)} \sin^{6}{\left(x \right)} \cos{\left(x \right)} d x} = \frac{2 \ln{\left(\sin{\left(x \right)} \right)} \sin^{7}{\left(x \right)}}{7} - \frac{2 \sin^{7}{\left(x \right)}}{49}$$
Sievennä:
$$\int{2 \ln{\left(\sin{\left(x \right)} \right)} \sin^{6}{\left(x \right)} \cos{\left(x \right)} d x} = \frac{2 \left(7 \ln{\left(\sin{\left(x \right)} \right)} - 1\right) \sin^{7}{\left(x \right)}}{49}$$
Lisää integrointivakio:
$$\int{2 \ln{\left(\sin{\left(x \right)} \right)} \sin^{6}{\left(x \right)} \cos{\left(x \right)} d x} = \frac{2 \left(7 \ln{\left(\sin{\left(x \right)} \right)} - 1\right) \sin^{7}{\left(x \right)}}{49}+C$$
Vastaus
$$$\int 2 \ln\left(\sin{\left(x \right)}\right) \sin^{6}{\left(x \right)} \cos{\left(x \right)}\, dx = \frac{2 \left(7 \ln\left(\sin{\left(x \right)}\right) - 1\right) \sin^{7}{\left(x \right)}}{49} + C$$$A