Intégrale de $\sin^{4}{\left(x \right)}$

Trouvez $\int \sin^{4}{\left(x \right)}\, dx$.

Rewrite the sine using the power reducing formula $\sin^{4}{\left(\alpha \right)} = - \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{\cos{\left(4 \alpha \right)}}{8} + \frac{3}{8}$ with $\alpha=x$:

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

Apply the constant multiple rule $\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$ with $c=\frac{1}{8}$ and $f{\left(x \right)} = - 4 \cos{\left(2 x \right)} + \cos{\left(4 x \right)} + 3$:

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

Integrate term by term:

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

Apply the constant rule $\int c\, dx = c x$ with $c=3$:

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

Apply the constant multiple rule $\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$ with $c=4$ and $f{\left(x \right)} = \cos{\left(2 x \right)}$:

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

Let $u=2 x$.

Then $du=\left(2 x\right)^{\prime }dx = 2 dx$ (steps can be seen »), and we have that $dx = \frac{du}{2}$.

C'est pourquoi,

$$\frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{2} = \frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{2}$$

Apply the constant multiple rule $\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$ with $c=\frac{1}{2}$ and $f{\left(u \right)} = \cos{\left(u \right)}$:

$$\frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{2} = \frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{2}$$

The integral of the cosine is $\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$:

$$\frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{4} = \frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\sin{\left(u \right)}}}}{4}$$

Recall that $u=2 x$:

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

Let $u=4 x$.

Then $du=\left(4 x\right)^{\prime }dx = 4 dx$ (steps can be seen »), and we have that $dx = \frac{du}{4}$.

The integral becomes

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

Apply the constant multiple rule $\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$ with $c=\frac{1}{4}$ and $f{\left(u \right)} = \cos{\left(u \right)}$:

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

The integral of the cosine is $\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$:

$$\frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{32} = \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\sin{\left(u \right)}}}}{32}$$

Recall that $u=4 x$:

$$\frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left({\color{red}{u}} \right)}}{32} = \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left({\color{red}{\left(4 x\right)}} \right)}}{32}$$

C'est pourquoi,

$$\int{\sin^{4}{\left(x \right)} d x} = \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{32}$$

Simplifier :

$$\int{\sin^{4}{\left(x \right)} d x} = \frac{12 x - 8 \sin{\left(2 x \right)} + \sin{\left(4 x \right)}}{32}$$

Ajouter la constante d'intégration :

$$\int{\sin^{4}{\left(x \right)} d x} = \frac{12 x - 8 \sin{\left(2 x \right)} + \sin{\left(4 x \right)}}{32}+C$$

Answer: $\int{\sin^{4}{\left(x \right)} d x}=\frac{12 x - 8 \sin{\left(2 x \right)} + \sin{\left(4 x \right)}}{32}+C$