Intégrale de $$$16 \sin{\left(\theta \right)} \cos^{4}{\left(\theta \right)}$$$

Déterminez $$$\int 16 \sin{\left(\theta \right)} \cos^{4}{\left(\theta \right)}\, d\theta$$$.

Appliquer la formule de réduction de puissance $$$\cos^{4}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{\cos{\left(4 \alpha \right)}}{8} + \frac{3}{8}$$$ avec $$$\alpha=\theta$$$:

$${\color{red}{\int{16 \sin{\left(\theta \right)} \cos^{4}{\left(\theta \right)} d \theta}}} = {\color{red}{\int{2 \left(4 \cos{\left(2 \theta \right)} + \cos{\left(4 \theta \right)} + 3\right) \sin{\left(\theta \right)} d \theta}}}$$

Appliquez la règle du facteur constant $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ avec $$$c=\frac{1}{8}$$$ et $$$f{\left(\theta \right)} = 16 \left(4 \cos{\left(2 \theta \right)} + \cos{\left(4 \theta \right)} + 3\right) \sin{\left(\theta \right)}$$$ :

$${\color{red}{\int{2 \left(4 \cos{\left(2 \theta \right)} + \cos{\left(4 \theta \right)} + 3\right) \sin{\left(\theta \right)} d \theta}}} = {\color{red}{\left(\frac{\int{16 \left(4 \cos{\left(2 \theta \right)} + \cos{\left(4 \theta \right)} + 3\right) \sin{\left(\theta \right)} d \theta}}{8}\right)}}$$

Expand the expression:

$$\frac{{\color{red}{\int{16 \left(4 \cos{\left(2 \theta \right)} + \cos{\left(4 \theta \right)} + 3\right) \sin{\left(\theta \right)} d \theta}}}}{8} = \frac{{\color{red}{\int{\left(64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} + 16 \sin{\left(\theta \right)} \cos{\left(4 \theta \right)} + 48 \sin{\left(\theta \right)}\right)d \theta}}}}{8}$$

Intégrez terme à terme:

$$\frac{{\color{red}{\int{\left(64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} + 16 \sin{\left(\theta \right)} \cos{\left(4 \theta \right)} + 48 \sin{\left(\theta \right)}\right)d \theta}}}}{8} = \frac{{\color{red}{\left(\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta} + \int{16 \sin{\left(\theta \right)} \cos{\left(4 \theta \right)} d \theta} + \int{48 \sin{\left(\theta \right)} d \theta}\right)}}}{8}$$

Appliquez la règle du facteur constant $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ avec $$$c=48$$$ et $$$f{\left(\theta \right)} = \sin{\left(\theta \right)}$$$ :

$$\frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{\int{16 \sin{\left(\theta \right)} \cos{\left(4 \theta \right)} d \theta}}{8} + \frac{{\color{red}{\int{48 \sin{\left(\theta \right)} d \theta}}}}{8} = \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{\int{16 \sin{\left(\theta \right)} \cos{\left(4 \theta \right)} d \theta}}{8} + \frac{{\color{red}{\left(48 \int{\sin{\left(\theta \right)} d \theta}\right)}}}{8}$$

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

$$\frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{\int{16 \sin{\left(\theta \right)} \cos{\left(4 \theta \right)} d \theta}}{8} + 6 {\color{red}{\int{\sin{\left(\theta \right)} d \theta}}} = \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{\int{16 \sin{\left(\theta \right)} \cos{\left(4 \theta \right)} d \theta}}{8} + 6 {\color{red}{\left(- \cos{\left(\theta \right)}\right)}}$$

Réécrivez $$$\sin\left(\theta \right)\cos\left(4 \theta \right)$$$ à l'aide de la formule $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$ avec $$$\alpha=\theta$$$ et $$$\beta=4 \theta$$$:

$$- 6 \cos{\left(\theta \right)} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{{\color{red}{\int{16 \sin{\left(\theta \right)} \cos{\left(4 \theta \right)} d \theta}}}}{8} = - 6 \cos{\left(\theta \right)} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{{\color{red}{\int{\left(- 8 \sin{\left(3 \theta \right)} + 8 \sin{\left(5 \theta \right)}\right)d \theta}}}}{8}$$

Appliquez la règle du facteur constant $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ avec $$$c=\frac{1}{2}$$$ et $$$f{\left(\theta \right)} = - 16 \sin{\left(3 \theta \right)} + 16 \sin{\left(5 \theta \right)}$$$ :

$$- 6 \cos{\left(\theta \right)} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{{\color{red}{\int{\left(- 8 \sin{\left(3 \theta \right)} + 8 \sin{\left(5 \theta \right)}\right)d \theta}}}}{8} = - 6 \cos{\left(\theta \right)} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{{\color{red}{\left(\frac{\int{\left(- 16 \sin{\left(3 \theta \right)} + 16 \sin{\left(5 \theta \right)}\right)d \theta}}{2}\right)}}}{8}$$

Intégrez terme à terme:

$$- 6 \cos{\left(\theta \right)} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{{\color{red}{\int{\left(- 16 \sin{\left(3 \theta \right)} + 16 \sin{\left(5 \theta \right)}\right)d \theta}}}}{16} = - 6 \cos{\left(\theta \right)} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{{\color{red}{\left(- \int{16 \sin{\left(3 \theta \right)} d \theta} + \int{16 \sin{\left(5 \theta \right)} d \theta}\right)}}}{16}$$

Appliquez la règle du facteur constant $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ avec $$$c=16$$$ et $$$f{\left(\theta \right)} = \sin{\left(3 \theta \right)}$$$ :

$$- 6 \cos{\left(\theta \right)} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{\int{16 \sin{\left(5 \theta \right)} d \theta}}{16} - \frac{{\color{red}{\int{16 \sin{\left(3 \theta \right)} d \theta}}}}{16} = - 6 \cos{\left(\theta \right)} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{\int{16 \sin{\left(5 \theta \right)} d \theta}}{16} - \frac{{\color{red}{\left(16 \int{\sin{\left(3 \theta \right)} d \theta}\right)}}}{16}$$

Soit $$$u=3 \theta$$$.

Alors $$$du=\left(3 \theta\right)^{\prime }d\theta = 3 d\theta$$$ (les étapes peuvent être vues »), et nous obtenons $$$d\theta = \frac{du}{3}$$$.

Par conséquent,

$$- 6 \cos{\left(\theta \right)} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{\int{16 \sin{\left(5 \theta \right)} d \theta}}{16} - {\color{red}{\int{\sin{\left(3 \theta \right)} d \theta}}} = - 6 \cos{\left(\theta \right)} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{\int{16 \sin{\left(5 \theta \right)} d \theta}}{16} - {\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}}$$

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 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ :

$$- 6 \cos{\left(\theta \right)} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{\int{16 \sin{\left(5 \theta \right)} d \theta}}{16} - {\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}} = - 6 \cos{\left(\theta \right)} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{\int{16 \sin{\left(5 \theta \right)} d \theta}}{16} - {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{3}\right)}}$$

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

$$- 6 \cos{\left(\theta \right)} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{\int{16 \sin{\left(5 \theta \right)} d \theta}}{16} - \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{3} = - 6 \cos{\left(\theta \right)} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{\int{16 \sin{\left(5 \theta \right)} d \theta}}{16} - \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{3}$$

Rappelons que $$$u=3 \theta$$$ :

$$- 6 \cos{\left(\theta \right)} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{\int{16 \sin{\left(5 \theta \right)} d \theta}}{16} + \frac{\cos{\left({\color{red}{u}} \right)}}{3} = - 6 \cos{\left(\theta \right)} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{\int{16 \sin{\left(5 \theta \right)} d \theta}}{16} + \frac{\cos{\left({\color{red}{\left(3 \theta\right)}} \right)}}{3}$$

Appliquez la règle du facteur constant $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ avec $$$c=16$$$ et $$$f{\left(\theta \right)} = \sin{\left(5 \theta \right)}$$$ :

$$- 6 \cos{\left(\theta \right)} + \frac{\cos{\left(3 \theta \right)}}{3} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{{\color{red}{\int{16 \sin{\left(5 \theta \right)} d \theta}}}}{16} = - 6 \cos{\left(\theta \right)} + \frac{\cos{\left(3 \theta \right)}}{3} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{{\color{red}{\left(16 \int{\sin{\left(5 \theta \right)} d \theta}\right)}}}{16}$$

Soit $$$u=5 \theta$$$.

Alors $$$du=\left(5 \theta\right)^{\prime }d\theta = 5 d\theta$$$ (les étapes peuvent être vues »), et nous obtenons $$$d\theta = \frac{du}{5}$$$.

L’intégrale devient

$$- 6 \cos{\left(\theta \right)} + \frac{\cos{\left(3 \theta \right)}}{3} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + {\color{red}{\int{\sin{\left(5 \theta \right)} d \theta}}} = - 6 \cos{\left(\theta \right)} + \frac{\cos{\left(3 \theta \right)}}{3} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + {\color{red}{\int{\frac{\sin{\left(u \right)}}{5} d u}}}$$

Appliquez la règle du facteur constant $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ avec $$$c=\frac{1}{5}$$$ et $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ :

$$- 6 \cos{\left(\theta \right)} + \frac{\cos{\left(3 \theta \right)}}{3} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + {\color{red}{\int{\frac{\sin{\left(u \right)}}{5} d u}}} = - 6 \cos{\left(\theta \right)} + \frac{\cos{\left(3 \theta \right)}}{3} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{5}\right)}}$$

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

$$- 6 \cos{\left(\theta \right)} + \frac{\cos{\left(3 \theta \right)}}{3} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{5} = - 6 \cos{\left(\theta \right)} + \frac{\cos{\left(3 \theta \right)}}{3} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{5}$$

Rappelons que $$$u=5 \theta$$$ :

$$- 6 \cos{\left(\theta \right)} + \frac{\cos{\left(3 \theta \right)}}{3} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} - \frac{\cos{\left({\color{red}{u}} \right)}}{5} = - 6 \cos{\left(\theta \right)} + \frac{\cos{\left(3 \theta \right)}}{3} + \frac{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}{8} - \frac{\cos{\left({\color{red}{\left(5 \theta\right)}} \right)}}{5}$$

Réécrivez $$$\sin\left(\theta \right)\cos\left(2 \theta \right)$$$ à l'aide de la formule $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$ avec $$$\alpha=\theta$$$ et $$$\beta=2 \theta$$$:

$$- 6 \cos{\left(\theta \right)} + \frac{\cos{\left(3 \theta \right)}}{3} - \frac{\cos{\left(5 \theta \right)}}{5} + \frac{{\color{red}{\int{64 \sin{\left(\theta \right)} \cos{\left(2 \theta \right)} d \theta}}}}{8} = - 6 \cos{\left(\theta \right)} + \frac{\cos{\left(3 \theta \right)}}{3} - \frac{\cos{\left(5 \theta \right)}}{5} + \frac{{\color{red}{\int{\left(- 32 \sin{\left(\theta \right)} + 32 \sin{\left(3 \theta \right)}\right)d \theta}}}}{8}$$

Appliquez la règle du facteur constant $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ avec $$$c=\frac{1}{2}$$$ et $$$f{\left(\theta \right)} = - 64 \sin{\left(\theta \right)} + 64 \sin{\left(3 \theta \right)}$$$ :

$$- 6 \cos{\left(\theta \right)} + \frac{\cos{\left(3 \theta \right)}}{3} - \frac{\cos{\left(5 \theta \right)}}{5} + \frac{{\color{red}{\int{\left(- 32 \sin{\left(\theta \right)} + 32 \sin{\left(3 \theta \right)}\right)d \theta}}}}{8} = - 6 \cos{\left(\theta \right)} + \frac{\cos{\left(3 \theta \right)}}{3} - \frac{\cos{\left(5 \theta \right)}}{5} + \frac{{\color{red}{\left(\frac{\int{\left(- 64 \sin{\left(\theta \right)} + 64 \sin{\left(3 \theta \right)}\right)d \theta}}{2}\right)}}}{8}$$

Intégrez terme à terme:

$$- 6 \cos{\left(\theta \right)} + \frac{\cos{\left(3 \theta \right)}}{3} - \frac{\cos{\left(5 \theta \right)}}{5} + \frac{{\color{red}{\int{\left(- 64 \sin{\left(\theta \right)} + 64 \sin{\left(3 \theta \right)}\right)d \theta}}}}{16} = - 6 \cos{\left(\theta \right)} + \frac{\cos{\left(3 \theta \right)}}{3} - \frac{\cos{\left(5 \theta \right)}}{5} + \frac{{\color{red}{\left(- \int{64 \sin{\left(\theta \right)} d \theta} + \int{64 \sin{\left(3 \theta \right)} d \theta}\right)}}}{16}$$

Appliquez la règle du facteur constant $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ avec $$$c=64$$$ et $$$f{\left(\theta \right)} = \sin{\left(\theta \right)}$$$ :

$$- 6 \cos{\left(\theta \right)} + \frac{\cos{\left(3 \theta \right)}}{3} - \frac{\cos{\left(5 \theta \right)}}{5} + \frac{\int{64 \sin{\left(3 \theta \right)} d \theta}}{16} - \frac{{\color{red}{\int{64 \sin{\left(\theta \right)} d \theta}}}}{16} = - 6 \cos{\left(\theta \right)} + \frac{\cos{\left(3 \theta \right)}}{3} - \frac{\cos{\left(5 \theta \right)}}{5} + \frac{\int{64 \sin{\left(3 \theta \right)} d \theta}}{16} - \frac{{\color{red}{\left(64 \int{\sin{\left(\theta \right)} d \theta}\right)}}}{16}$$

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

$$- 6 \cos{\left(\theta \right)} + \frac{\cos{\left(3 \theta \right)}}{3} - \frac{\cos{\left(5 \theta \right)}}{5} + \frac{\int{64 \sin{\left(3 \theta \right)} d \theta}}{16} - 4 {\color{red}{\int{\sin{\left(\theta \right)} d \theta}}} = - 6 \cos{\left(\theta \right)} + \frac{\cos{\left(3 \theta \right)}}{3} - \frac{\cos{\left(5 \theta \right)}}{5} + \frac{\int{64 \sin{\left(3 \theta \right)} d \theta}}{16} - 4 {\color{red}{\left(- \cos{\left(\theta \right)}\right)}}$$

Appliquez la règle du facteur constant $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ avec $$$c=64$$$ et $$$f{\left(\theta \right)} = \sin{\left(3 \theta \right)}$$$ :

$$- 2 \cos{\left(\theta \right)} + \frac{\cos{\left(3 \theta \right)}}{3} - \frac{\cos{\left(5 \theta \right)}}{5} + \frac{{\color{red}{\int{64 \sin{\left(3 \theta \right)} d \theta}}}}{16} = - 2 \cos{\left(\theta \right)} + \frac{\cos{\left(3 \theta \right)}}{3} - \frac{\cos{\left(5 \theta \right)}}{5} + \frac{{\color{red}{\left(64 \int{\sin{\left(3 \theta \right)} d \theta}\right)}}}{16}$$

L'intégrale $$$\int{\sin{\left(3 \theta \right)} d \theta}$$$ a déjà été calculée :

$$\int{\sin{\left(3 \theta \right)} d \theta} = - \frac{\cos{\left(3 \theta \right)}}{3}$$

Par conséquent,

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

Par conséquent,

$$\int{16 \sin{\left(\theta \right)} \cos^{4}{\left(\theta \right)} d \theta} = - 2 \cos{\left(\theta \right)} - \cos{\left(3 \theta \right)} - \frac{\cos{\left(5 \theta \right)}}{5}$$

Ajouter la constante d'intégration :

$$\int{16 \sin{\left(\theta \right)} \cos^{4}{\left(\theta \right)} d \theta} = - 2 \cos{\left(\theta \right)} - \cos{\left(3 \theta \right)} - \frac{\cos{\left(5 \theta \right)}}{5}+C$$

$$$\int 16 \sin{\left(\theta \right)} \cos^{4}{\left(\theta \right)}\, d\theta = \left(- 2 \cos{\left(\theta \right)} - \cos{\left(3 \theta \right)} - \frac{\cos{\left(5 \theta \right)}}{5}\right) + C$$$A