Intégrale de $$$\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} - 4 \sin{\left(x \right)} - \cos^{2}{\left(x \right)}$$$

Déterminez $$$\int \left(\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} - 4 \sin{\left(x \right)} - \cos^{2}{\left(x \right)}\right)\, dx$$$.

Intégrez terme à terme:

$${\color{red}{\int{\left(\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} - 4 \sin{\left(x \right)} - \cos^{2}{\left(x \right)}\right)d x}}} = {\color{red}{\left(\int{\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} d x} - \int{4 \sin{\left(x \right)} d x} - \int{\cos^{2}{\left(x \right)} d x}\right)}}$$

Appliquer la formule de réduction de puissance $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$ avec $$$\alpha=x$$$:

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

Appliquez la règle du facteur constant $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ avec $$$c=\frac{1}{2}$$$ et $$$f{\left(x \right)} = \cos{\left(2 x \right)} + 1$$$ :

$$\int{\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} d x} - \int{4 \sin{\left(x \right)} d x} - {\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)d x}}} = \int{\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} d x} - \int{4 \sin{\left(x \right)} d x} - {\color{red}{\left(\frac{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}{2}\right)}}$$

Intégrez terme à terme:

$$\int{\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} d x} - \int{4 \sin{\left(x \right)} d x} - \frac{{\color{red}{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}}}{2} = \int{\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} d x} - \int{4 \sin{\left(x \right)} d x} - \frac{{\color{red}{\left(\int{1 d x} + \int{\cos{\left(2 x \right)} d x}\right)}}}{2}$$

Appliquez la règle de la constante $$$\int c\, dx = c x$$$ avec $$$c=1$$$:

$$\int{\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} d x} - \int{4 \sin{\left(x \right)} d x} - \frac{\int{\cos{\left(2 x \right)} d x}}{2} - \frac{{\color{red}{\int{1 d x}}}}{2} = \int{\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} d x} - \int{4 \sin{\left(x \right)} d x} - \frac{\int{\cos{\left(2 x \right)} d x}}{2} - \frac{{\color{red}{x}}}{2}$$

Soit $$$u=2 x$$$.

Alors $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (les étapes peuvent être vues »), et nous obtenons $$$dx = \frac{du}{2}$$$.

Donc,

$$- \frac{x}{2} + \int{\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} d x} - \int{4 \sin{\left(x \right)} d x} - \frac{{\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{2} = - \frac{x}{2} + \int{\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} d x} - \int{4 \sin{\left(x \right)} d x} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{2}$$

Appliquez la règle du facteur constant $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ avec $$$c=\frac{1}{2}$$$ et $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ :

$$- \frac{x}{2} + \int{\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} d x} - \int{4 \sin{\left(x \right)} d x} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{2} = - \frac{x}{2} + \int{\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} d x} - \int{4 \sin{\left(x \right)} d x} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{2}$$

L’intégrale du cosinus est $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$ :

$$- \frac{x}{2} + \int{\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} d x} - \int{4 \sin{\left(x \right)} d x} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{4} = - \frac{x}{2} + \int{\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} d x} - \int{4 \sin{\left(x \right)} d x} - \frac{{\color{red}{\sin{\left(u \right)}}}}{4}$$

Rappelons que $$$u=2 x$$$ :

$$- \frac{x}{2} + \int{\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} d x} - \int{4 \sin{\left(x \right)} d x} - \frac{\sin{\left({\color{red}{u}} \right)}}{4} = - \frac{x}{2} + \int{\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} d x} - \int{4 \sin{\left(x \right)} d x} - \frac{\sin{\left({\color{red}{\left(2 x\right)}} \right)}}{4}$$

Appliquez la règle du facteur constant $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ avec $$$c=4$$$ et $$$f{\left(x \right)} = \sin{\left(x \right)}$$$ :

$$- \frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4} + \int{\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} d x} - {\color{red}{\int{4 \sin{\left(x \right)} d x}}} = - \frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4} + \int{\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} d x} - {\color{red}{\left(4 \int{\sin{\left(x \right)} d x}\right)}}$$

L’intégrale du sinus est $$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$ :

$$- \frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4} + \int{\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} d x} - 4 {\color{red}{\int{\sin{\left(x \right)} d x}}} = - \frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4} + \int{\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} d x} - 4 {\color{red}{\left(- \cos{\left(x \right)}\right)}}$$

Appliquez la règle du facteur constant $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ avec $$$c=\frac{1}{5}$$$ et $$$f{\left(x \right)} = \left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}$$$ :

$$- \frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4} + 4 \cos{\left(x \right)} + {\color{red}{\int{\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} d x}}} = - \frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4} + 4 \cos{\left(x \right)} + {\color{red}{\left(\frac{\int{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)} d x}}{5}\right)}}$$

Soit $$$u=3 \sin{\left(x \right)} - 2$$$.

Alors $$$du=\left(3 \sin{\left(x \right)} - 2\right)^{\prime }dx = 3 \cos{\left(x \right)} dx$$$ (les étapes peuvent être vues »), et nous obtenons $$$\cos{\left(x \right)} dx = \frac{du}{3}$$$.

Donc,

$$- \frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4} + 4 \cos{\left(x \right)} + \frac{{\color{red}{\int{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)} d x}}}}{5} = - \frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4} + 4 \cos{\left(x \right)} + \frac{{\color{red}{\int{\frac{u}{3} d u}}}}{5}$$

Appliquez la règle du facteur constant $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ avec $$$c=\frac{1}{3}$$$ et $$$f{\left(u \right)} = u$$$ :

$$- \frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4} + 4 \cos{\left(x \right)} + \frac{{\color{red}{\int{\frac{u}{3} d u}}}}{5} = - \frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4} + 4 \cos{\left(x \right)} + \frac{{\color{red}{\left(\frac{\int{u d u}}{3}\right)}}}{5}$$

Appliquer la règle de puissance $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ avec $$$n=1$$$ :

$$- \frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4} + 4 \cos{\left(x \right)} + \frac{{\color{red}{\int{u d u}}}}{15}=- \frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4} + 4 \cos{\left(x \right)} + \frac{{\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{15}=- \frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4} + 4 \cos{\left(x \right)} + \frac{{\color{red}{\left(\frac{u^{2}}{2}\right)}}}{15}$$

Rappelons que $$$u=3 \sin{\left(x \right)} - 2$$$ :

$$- \frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4} + 4 \cos{\left(x \right)} + \frac{{\color{red}{u}}^{2}}{30} = - \frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4} + 4 \cos{\left(x \right)} + \frac{{\color{red}{\left(3 \sin{\left(x \right)} - 2\right)}}^{2}}{30}$$

Par conséquent,

$$\int{\left(\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} - 4 \sin{\left(x \right)} - \cos^{2}{\left(x \right)}\right)d x} = - \frac{x}{2} + \frac{\left(3 \sin{\left(x \right)} - 2\right)^{2}}{30} - \frac{\sin{\left(2 x \right)}}{4} + 4 \cos{\left(x \right)}$$

Ajouter la constante d'intégration :

$$\int{\left(\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} - 4 \sin{\left(x \right)} - \cos^{2}{\left(x \right)}\right)d x} = - \frac{x}{2} + \frac{\left(3 \sin{\left(x \right)} - 2\right)^{2}}{30} - \frac{\sin{\left(2 x \right)}}{4} + 4 \cos{\left(x \right)}+C$$

$$$\int \left(\frac{\left(3 \sin{\left(x \right)} - 2\right) \cos{\left(x \right)}}{5} - 4 \sin{\left(x \right)} - \cos^{2}{\left(x \right)}\right)\, dx = \left(- \frac{x}{2} + \frac{\left(3 \sin{\left(x \right)} - 2\right)^{2}}{30} - \frac{\sin{\left(2 x \right)}}{4} + 4 \cos{\left(x \right)}\right) + C$$$A