Integrale di $$$\sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)}$$$

Trova $$$\int \sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)}\, dx$$$.

Estrai un coseno e scrivi tutto il resto in termini del seno, usando la formula $$$\cos^2\left(\alpha \right)=-\sin^2\left(\alpha \right)+1$$$ con $$$\alpha=x$$$:

$${\color{red}{\int{\sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)} d x}}} = {\color{red}{\int{\left(1 - \sin^{2}{\left(x \right)}\right)^{2} \sin^{2}{\left(x \right)} \cos{\left(x \right)} d x}}}$$

Sia $$$u=\sin{\left(x \right)}$$$.

Quindi $$$du=\left(\sin{\left(x \right)}\right)^{\prime }dx = \cos{\left(x \right)} dx$$$ (i passaggi si possono vedere »), e si ha che $$$\cos{\left(x \right)} dx = du$$$.

L'integrale diventa

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

Expand the expression:

$${\color{red}{\int{u^{2} \left(1 - u^{2}\right)^{2} d u}}} = {\color{red}{\int{\left(u^{6} - 2 u^{4} + u^{2}\right)d u}}}$$

Integra termine per termine:

$${\color{red}{\int{\left(u^{6} - 2 u^{4} + u^{2}\right)d u}}} = {\color{red}{\left(\int{u^{2} d u} - \int{2 u^{4} d u} + \int{u^{6} d u}\right)}}$$

Applica la regola della potenza $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=2$$$:

$$- \int{2 u^{4} d u} + \int{u^{6} d u} + {\color{red}{\int{u^{2} d u}}}=- \int{2 u^{4} d u} + \int{u^{6} d u} + {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=- \int{2 u^{4} d u} + \int{u^{6} d u} + {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

Applica la regola della potenza $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=6$$$:

$$\frac{u^{3}}{3} - \int{2 u^{4} d u} + {\color{red}{\int{u^{6} d u}}}=\frac{u^{3}}{3} - \int{2 u^{4} d u} + {\color{red}{\frac{u^{1 + 6}}{1 + 6}}}=\frac{u^{3}}{3} - \int{2 u^{4} d u} + {\color{red}{\left(\frac{u^{7}}{7}\right)}}$$

Applica la regola del fattore costante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=2$$$ e $$$f{\left(u \right)} = u^{4}$$$:

$$\frac{u^{7}}{7} + \frac{u^{3}}{3} - {\color{red}{\int{2 u^{4} d u}}} = \frac{u^{7}}{7} + \frac{u^{3}}{3} - {\color{red}{\left(2 \int{u^{4} d u}\right)}}$$

Applica la regola della potenza $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=4$$$:

$$\frac{u^{7}}{7} + \frac{u^{3}}{3} - 2 {\color{red}{\int{u^{4} d u}}}=\frac{u^{7}}{7} + \frac{u^{3}}{3} - 2 {\color{red}{\frac{u^{1 + 4}}{1 + 4}}}=\frac{u^{7}}{7} + \frac{u^{3}}{3} - 2 {\color{red}{\left(\frac{u^{5}}{5}\right)}}$$

Ricordiamo che $$$u=\sin{\left(x \right)}$$$:

$$\frac{{\color{red}{u}}^{3}}{3} - \frac{2 {\color{red}{u}}^{5}}{5} + \frac{{\color{red}{u}}^{7}}{7} = \frac{{\color{red}{\sin{\left(x \right)}}}^{3}}{3} - \frac{2 {\color{red}{\sin{\left(x \right)}}}^{5}}{5} + \frac{{\color{red}{\sin{\left(x \right)}}}^{7}}{7}$$

Pertanto,

$$\int{\sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)} d x} = \frac{\sin^{7}{\left(x \right)}}{7} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{\sin^{3}{\left(x \right)}}{3}$$

Semplifica:

$$\int{\sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)} d x} = \frac{\left(15 \sin^{4}{\left(x \right)} - 42 \sin^{2}{\left(x \right)} + 35\right) \sin^{3}{\left(x \right)}}{105}$$

Aggiungi la costante di integrazione:

$$\int{\sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)} d x} = \frac{\left(15 \sin^{4}{\left(x \right)} - 42 \sin^{2}{\left(x \right)} + 35\right) \sin^{3}{\left(x \right)}}{105}+C$$

$$$\int \sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)}\, dx = \frac{\left(15 \sin^{4}{\left(x \right)} - 42 \sin^{2}{\left(x \right)} + 35\right) \sin^{3}{\left(x \right)}}{105} + C$$$A