Intégrale de $$$2 \cos^{8}{\left(x \right)} - 2 \cos^{5}{\left(x \right)}$$$
Calculatrice associée: Calculatrice d’intégrales définies et impropres
Votre saisie
Déterminez $$$\int \left(2 \cos^{8}{\left(x \right)} - 2 \cos^{5}{\left(x \right)}\right)\, dx$$$.
Solution
Intégrez terme à terme:
$${\color{red}{\int{\left(2 \cos^{8}{\left(x \right)} - 2 \cos^{5}{\left(x \right)}\right)d x}}} = {\color{red}{\left(- \int{2 \cos^{5}{\left(x \right)} d x} + \int{2 \cos^{8}{\left(x \right)} d x}\right)}}$$
Appliquez la règle du facteur constant $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ avec $$$c=2$$$ et $$$f{\left(x \right)} = \cos^{5}{\left(x \right)}$$$ :
$$\int{2 \cos^{8}{\left(x \right)} d x} - {\color{red}{\int{2 \cos^{5}{\left(x \right)} d x}}} = \int{2 \cos^{8}{\left(x \right)} d x} - {\color{red}{\left(2 \int{\cos^{5}{\left(x \right)} d x}\right)}}$$
Isolez un cosinus et exprimez tout le reste en fonction du sinus, en utilisant la formule $$$\cos^2\left(\alpha \right)=-\sin^2\left(\alpha \right)+1$$$ avec $$$\alpha=x$$$:
$$\int{2 \cos^{8}{\left(x \right)} d x} - 2 {\color{red}{\int{\cos^{5}{\left(x \right)} d x}}} = \int{2 \cos^{8}{\left(x \right)} d x} - 2 {\color{red}{\int{\left(1 - \sin^{2}{\left(x \right)}\right)^{2} \cos{\left(x \right)} d x}}}$$
Soit $$$u=\sin{\left(x \right)}$$$.
Alors $$$du=\left(\sin{\left(x \right)}\right)^{\prime }dx = \cos{\left(x \right)} dx$$$ (les étapes peuvent être vues »), et nous obtenons $$$\cos{\left(x \right)} dx = du$$$.
Donc,
$$\int{2 \cos^{8}{\left(x \right)} d x} - 2 {\color{red}{\int{\left(1 - \sin^{2}{\left(x \right)}\right)^{2} \cos{\left(x \right)} d x}}} = \int{2 \cos^{8}{\left(x \right)} d x} - 2 {\color{red}{\int{\left(1 - u^{2}\right)^{2} d u}}}$$
Expand the expression:
$$\int{2 \cos^{8}{\left(x \right)} d x} - 2 {\color{red}{\int{\left(1 - u^{2}\right)^{2} d u}}} = \int{2 \cos^{8}{\left(x \right)} d x} - 2 {\color{red}{\int{\left(u^{4} - 2 u^{2} + 1\right)d u}}}$$
Intégrez terme à terme:
$$\int{2 \cos^{8}{\left(x \right)} d x} - 2 {\color{red}{\int{\left(u^{4} - 2 u^{2} + 1\right)d u}}} = \int{2 \cos^{8}{\left(x \right)} d x} - 2 {\color{red}{\left(\int{1 d u} - \int{2 u^{2} d u} + \int{u^{4} d u}\right)}}$$
Appliquez la règle de la constante $$$\int c\, du = c u$$$ avec $$$c=1$$$:
$$\int{2 \cos^{8}{\left(x \right)} d x} + 2 \int{2 u^{2} d u} - 2 \int{u^{4} d u} - 2 {\color{red}{\int{1 d u}}} = \int{2 \cos^{8}{\left(x \right)} d x} + 2 \int{2 u^{2} d u} - 2 \int{u^{4} d u} - 2 {\color{red}{u}}$$
Appliquer la règle de puissance $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ avec $$$n=4$$$ :
$$- 2 u + \int{2 \cos^{8}{\left(x \right)} d x} + 2 \int{2 u^{2} d u} - 2 {\color{red}{\int{u^{4} d u}}}=- 2 u + \int{2 \cos^{8}{\left(x \right)} d x} + 2 \int{2 u^{2} d u} - 2 {\color{red}{\frac{u^{1 + 4}}{1 + 4}}}=- 2 u + \int{2 \cos^{8}{\left(x \right)} d x} + 2 \int{2 u^{2} d u} - 2 {\color{red}{\left(\frac{u^{5}}{5}\right)}}$$
Appliquez la règle du facteur constant $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ avec $$$c=2$$$ et $$$f{\left(u \right)} = u^{2}$$$ :
$$- \frac{2 u^{5}}{5} - 2 u + \int{2 \cos^{8}{\left(x \right)} d x} + 2 {\color{red}{\int{2 u^{2} d u}}} = - \frac{2 u^{5}}{5} - 2 u + \int{2 \cos^{8}{\left(x \right)} d x} + 2 {\color{red}{\left(2 \int{u^{2} d u}\right)}}$$
Appliquer la règle de puissance $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ avec $$$n=2$$$ :
$$- \frac{2 u^{5}}{5} - 2 u + \int{2 \cos^{8}{\left(x \right)} d x} + 4 {\color{red}{\int{u^{2} d u}}}=- \frac{2 u^{5}}{5} - 2 u + \int{2 \cos^{8}{\left(x \right)} d x} + 4 {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=- \frac{2 u^{5}}{5} - 2 u + \int{2 \cos^{8}{\left(x \right)} d x} + 4 {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$
Rappelons que $$$u=\sin{\left(x \right)}$$$ :
$$\int{2 \cos^{8}{\left(x \right)} d x} - 2 {\color{red}{u}} + \frac{4 {\color{red}{u}}^{3}}{3} - \frac{2 {\color{red}{u}}^{5}}{5} = \int{2 \cos^{8}{\left(x \right)} d x} - 2 {\color{red}{\sin{\left(x \right)}}} + \frac{4 {\color{red}{\sin{\left(x \right)}}}^{3}}{3} - \frac{2 {\color{red}{\sin{\left(x \right)}}}^{5}}{5}$$
Appliquez la règle du facteur constant $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ avec $$$c=2$$$ et $$$f{\left(x \right)} = \cos^{8}{\left(x \right)}$$$ :
$$- \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + {\color{red}{\int{2 \cos^{8}{\left(x \right)} d x}}} = - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + {\color{red}{\left(2 \int{\cos^{8}{\left(x \right)} d x}\right)}}$$
Appliquer la formule de réduction de puissance $$$\cos^{8}{\left(\alpha \right)} = \frac{7 \cos{\left(2 \alpha \right)}}{16} + \frac{7 \cos{\left(4 \alpha \right)}}{32} + \frac{\cos{\left(6 \alpha \right)}}{16} + \frac{\cos{\left(8 \alpha \right)}}{128} + \frac{35}{128}$$$ avec $$$\alpha=x$$$:
$$- \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + 2 {\color{red}{\int{\cos^{8}{\left(x \right)} d x}}} = - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + 2 {\color{red}{\int{\left(\frac{7 \cos{\left(2 x \right)}}{16} + \frac{7 \cos{\left(4 x \right)}}{32} + \frac{\cos{\left(6 x \right)}}{16} + \frac{\cos{\left(8 x \right)}}{128} + \frac{35}{128}\right)d x}}}$$
Appliquez la règle du facteur constant $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ avec $$$c=\frac{1}{128}$$$ et $$$f{\left(x \right)} = 56 \cos{\left(2 x \right)} + 28 \cos{\left(4 x \right)} + 8 \cos{\left(6 x \right)} + \cos{\left(8 x \right)} + 35$$$ :
$$- \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + 2 {\color{red}{\int{\left(\frac{7 \cos{\left(2 x \right)}}{16} + \frac{7 \cos{\left(4 x \right)}}{32} + \frac{\cos{\left(6 x \right)}}{16} + \frac{\cos{\left(8 x \right)}}{128} + \frac{35}{128}\right)d x}}} = - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + 2 {\color{red}{\left(\frac{\int{\left(56 \cos{\left(2 x \right)} + 28 \cos{\left(4 x \right)} + 8 \cos{\left(6 x \right)} + \cos{\left(8 x \right)} + 35\right)d x}}{128}\right)}}$$
Intégrez terme à terme:
$$- \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{{\color{red}{\int{\left(56 \cos{\left(2 x \right)} + 28 \cos{\left(4 x \right)} + 8 \cos{\left(6 x \right)} + \cos{\left(8 x \right)} + 35\right)d x}}}}{64} = - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{{\color{red}{\left(\int{35 d x} + \int{56 \cos{\left(2 x \right)} d x} + \int{28 \cos{\left(4 x \right)} d x} + \int{8 \cos{\left(6 x \right)} d x} + \int{\cos{\left(8 x \right)} d x}\right)}}}{64}$$
Appliquez la règle de la constante $$$\int c\, dx = c x$$$ avec $$$c=35$$$:
$$- \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{8 \cos{\left(6 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\int{35 d x}}}}{64} = - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{8 \cos{\left(6 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\left(35 x\right)}}}{64}$$
Appliquez la règle du facteur constant $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ avec $$$c=8$$$ et $$$f{\left(x \right)} = \cos{\left(6 x \right)}$$$ :
$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\int{8 \cos{\left(6 x \right)} d x}}}}{64} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\left(8 \int{\cos{\left(6 x \right)} d x}\right)}}}{64}$$
Soit $$$u=6 x$$$.
Alors $$$du=\left(6 x\right)^{\prime }dx = 6 dx$$$ (les étapes peuvent être vues »), et nous obtenons $$$dx = \frac{du}{6}$$$.
L’intégrale peut être réécrite sous la forme
$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\int{\cos{\left(6 x \right)} d x}}}}{8} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{6} d u}}}}{8}$$
Appliquez la règle du facteur constant $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ avec $$$c=\frac{1}{6}$$$ et $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ :
$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{6} d u}}}}{8} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{6}\right)}}}{8}$$
L’intégrale du cosinus est $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$ :
$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{48} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\sin{\left(u \right)}}}}{48}$$
Rappelons que $$$u=6 x$$$ :
$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{\sin{\left({\color{red}{u}} \right)}}{48} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{28 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{\sin{\left({\color{red}{\left(6 x\right)}} \right)}}{48}$$
Appliquez la règle du facteur constant $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ avec $$$c=28$$$ et $$$f{\left(x \right)} = \cos{\left(4 x \right)}$$$ :
$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\int{28 \cos{\left(4 x \right)} d x}}}}{64} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\left(28 \int{\cos{\left(4 x \right)} d x}\right)}}}{64}$$
Soit $$$u=4 x$$$.
Alors $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (les étapes peuvent être vues »), et nous obtenons $$$dx = \frac{du}{4}$$$.
L’intégrale peut être réécrite sous la forme
$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\int{\cos{\left(4 x \right)} d x}}}}{16} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{16}$$
Appliquez la règle du facteur constant $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ avec $$$c=\frac{1}{4}$$$ et $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ :
$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{16} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}}{16}$$
L’intégrale du cosinus est $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$ :
$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\int{\cos{\left(u \right)} d u}}}}{64} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\sin{\left(u \right)}}}}{64}$$
Rappelons que $$$u=4 x$$$ :
$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 \sin{\left({\color{red}{u}} \right)}}{64} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{56 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 \sin{\left({\color{red}{\left(4 x\right)}} \right)}}{64}$$
Appliquez la règle du facteur constant $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ avec $$$c=56$$$ et $$$f{\left(x \right)} = \cos{\left(2 x \right)}$$$ :
$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\int{56 \cos{\left(2 x \right)} d x}}}}{64} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{{\color{red}{\left(56 \int{\cos{\left(2 x \right)} d x}\right)}}}{64}$$
Soit $$$u=2 x$$$.
Alors $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (les étapes peuvent être vues »), et nous obtenons $$$dx = \frac{du}{2}$$$.
Par conséquent,
$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{8} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{8}$$
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 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ :
$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{8} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{8}$$
L’intégrale du cosinus est $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$ :
$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\int{\cos{\left(u \right)} d u}}}}{16} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 {\color{red}{\sin{\left(u \right)}}}}{16}$$
Rappelons que $$$u=2 x$$$ :
$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 \sin{\left({\color{red}{u}} \right)}}{16} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{64} + \frac{7 \sin{\left({\color{red}{\left(2 x\right)}} \right)}}{16}$$
Soit $$$u=8 x$$$.
Alors $$$du=\left(8 x\right)^{\prime }dx = 8 dx$$$ (les étapes peuvent être vues »), et nous obtenons $$$dx = \frac{du}{8}$$$.
Ainsi,
$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(2 x \right)}}{16} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{{\color{red}{\int{\cos{\left(8 x \right)} d x}}}}{64} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(2 x \right)}}{16} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{8} d u}}}}{64}$$
Appliquez la règle du facteur constant $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ avec $$$c=\frac{1}{8}$$$ et $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ :
$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(2 x \right)}}{16} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{8} d u}}}}{64} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(2 x \right)}}{16} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{8}\right)}}}{64}$$
L’intégrale du cosinus est $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$ :
$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(2 x \right)}}{16} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{512} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(2 x \right)}}{16} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{{\color{red}{\sin{\left(u \right)}}}}{512}$$
Rappelons que $$$u=8 x$$$ :
$$\frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(2 x \right)}}{16} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\sin{\left({\color{red}{u}} \right)}}{512} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(2 x \right)}}{16} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\sin{\left({\color{red}{\left(8 x\right)}} \right)}}{512}$$
Par conséquent,
$$\int{\left(2 \cos^{8}{\left(x \right)} - 2 \cos^{5}{\left(x \right)}\right)d x} = \frac{35 x}{64} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3} - 2 \sin{\left(x \right)} + \frac{7 \sin{\left(2 x \right)}}{16} + \frac{7 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{48} + \frac{\sin{\left(8 x \right)}}{512}$$
Simplifier:
$$\int{\left(2 \cos^{8}{\left(x \right)} - 2 \cos^{5}{\left(x \right)}\right)d x} = \frac{4200 x - 3072 \sin^{5}{\left(x \right)} + 10240 \sin^{3}{\left(x \right)} - 15360 \sin{\left(x \right)} + 3360 \sin{\left(2 x \right)} + 840 \sin{\left(4 x \right)} + 160 \sin{\left(6 x \right)} + 15 \sin{\left(8 x \right)}}{7680}$$
Ajouter la constante d'intégration :
$$\int{\left(2 \cos^{8}{\left(x \right)} - 2 \cos^{5}{\left(x \right)}\right)d x} = \frac{4200 x - 3072 \sin^{5}{\left(x \right)} + 10240 \sin^{3}{\left(x \right)} - 15360 \sin{\left(x \right)} + 3360 \sin{\left(2 x \right)} + 840 \sin{\left(4 x \right)} + 160 \sin{\left(6 x \right)} + 15 \sin{\left(8 x \right)}}{7680}+C$$
Réponse
$$$\int \left(2 \cos^{8}{\left(x \right)} - 2 \cos^{5}{\left(x \right)}\right)\, dx = \frac{4200 x - 3072 \sin^{5}{\left(x \right)} + 10240 \sin^{3}{\left(x \right)} - 15360 \sin{\left(x \right)} + 3360 \sin{\left(2 x \right)} + 840 \sin{\left(4 x \right)} + 160 \sin{\left(6 x \right)} + 15 \sin{\left(8 x \right)}}{7680} + C$$$A