Integral of $\sin^{6}{\left(x \right)}$

Find $\int \sin^{6}{\left(x \right)}\, dx$.

Rewrite the sine using the power reducing formula $\sin^{6}{\left(\alpha \right)} = - \frac{15 \cos{\left(2 \alpha \right)}}{32} + \frac{3 \cos{\left(4 \alpha \right)}}{16} - \frac{\cos{\left(6 \alpha \right)}}{32} + \frac{5}{16}$ with $\alpha=x$:

$${\color{red}{\int{\sin^{6}{\left(x \right)} d x}}} = {\color{red}{\int{\left(- \frac{15 \cos{\left(2 x \right)}}{32} + \frac{3 \cos{\left(4 x \right)}}{16} - \frac{\cos{\left(6 x \right)}}{32} + \frac{5}{16}\right)d x}}}$$

Apply the constant multiple rule $\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$ with $c=\frac{1}{32}$ and $f{\left(x \right)} = - 15 \cos{\left(2 x \right)} + 6 \cos{\left(4 x \right)} - \cos{\left(6 x \right)} + 10$:

$${\color{red}{\int{\left(- \frac{15 \cos{\left(2 x \right)}}{32} + \frac{3 \cos{\left(4 x \right)}}{16} - \frac{\cos{\left(6 x \right)}}{32} + \frac{5}{16}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(- 15 \cos{\left(2 x \right)} + 6 \cos{\left(4 x \right)} - \cos{\left(6 x \right)} + 10\right)d x}}{32}\right)}}$$

Integrate term by term:

$$\frac{{\color{red}{\int{\left(- 15 \cos{\left(2 x \right)} + 6 \cos{\left(4 x \right)} - \cos{\left(6 x \right)} + 10\right)d x}}}}{32} = \frac{{\color{red}{\left(\int{10 d x} - \int{15 \cos{\left(2 x \right)} d x} + \int{6 \cos{\left(4 x \right)} d x} - \int{\cos{\left(6 x \right)} d x}\right)}}}{32}$$

Apply the constant rule $\int c\, dx = c x$ with $c=10$:

$$- \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{\int{\cos{\left(6 x \right)} d x}}{32} + \frac{{\color{red}{\int{10 d x}}}}{32} = - \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{\int{\cos{\left(6 x \right)} d x}}{32} + \frac{{\color{red}{\left(10 x\right)}}}{32}$$

Let $u=6 x$.

Then $du=\left(6 x\right)^{\prime }dx = 6 dx$ (steps can be seen »), and we have that $dx = \frac{du}{6}$.

The integral can be rewritten as

$$\frac{5 x}{16} - \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{{\color{red}{\int{\cos{\left(6 x \right)} d x}}}}{32} = \frac{5 x}{16} - \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{6} d u}}}}{32}$$

Apply the constant multiple rule $\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$ with $c=\frac{1}{6}$ and $f{\left(u \right)} = \cos{\left(u \right)}$:

$$\frac{5 x}{16} - \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{6} d u}}}}{32} = \frac{5 x}{16} - \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{6}\right)}}}{32}$$

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

$$\frac{5 x}{16} - \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{192} = \frac{5 x}{16} - \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{{\color{red}{\sin{\left(u \right)}}}}{192}$$

Recall that $u=6 x$:

$$\frac{5 x}{16} - \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{\sin{\left({\color{red}{u}} \right)}}{192} = \frac{5 x}{16} - \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{\sin{\left({\color{red}{\left(6 x\right)}} \right)}}{192}$$

Apply the constant multiple rule $\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$ with $c=15$ and $f{\left(x \right)} = \cos{\left(2 x \right)}$:

$$\frac{5 x}{16} - \frac{\sin{\left(6 x \right)}}{192} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{{\color{red}{\int{15 \cos{\left(2 x \right)} d x}}}}{32} = \frac{5 x}{16} - \frac{\sin{\left(6 x \right)}}{192} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{{\color{red}{\left(15 \int{\cos{\left(2 x \right)} d x}\right)}}}{32}$$

Let $u=2 x$.

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

Therefore,

$$\frac{5 x}{16} - \frac{\sin{\left(6 x \right)}}{192} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{15 {\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{32} = \frac{5 x}{16} - \frac{\sin{\left(6 x \right)}}{192} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{15 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{32}$$

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

$$\frac{5 x}{16} - \frac{\sin{\left(6 x \right)}}{192} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{15 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{32} = \frac{5 x}{16} - \frac{\sin{\left(6 x \right)}}{192} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{15 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{32}$$

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

$$\frac{5 x}{16} - \frac{\sin{\left(6 x \right)}}{192} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{15 {\color{red}{\int{\cos{\left(u \right)} d u}}}}{64} = \frac{5 x}{16} - \frac{\sin{\left(6 x \right)}}{192} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{15 {\color{red}{\sin{\left(u \right)}}}}{64}$$

Recall that $u=2 x$:

$$\frac{5 x}{16} - \frac{\sin{\left(6 x \right)}}{192} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{15 \sin{\left({\color{red}{u}} \right)}}{64} = \frac{5 x}{16} - \frac{\sin{\left(6 x \right)}}{192} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{15 \sin{\left({\color{red}{\left(2 x\right)}} \right)}}{64}$$

Apply the constant multiple rule $\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$ with $c=6$ and $f{\left(x \right)} = \cos{\left(4 x \right)}$:

$$\frac{5 x}{16} - \frac{15 \sin{\left(2 x \right)}}{64} - \frac{\sin{\left(6 x \right)}}{192} + \frac{{\color{red}{\int{6 \cos{\left(4 x \right)} d x}}}}{32} = \frac{5 x}{16} - \frac{15 \sin{\left(2 x \right)}}{64} - \frac{\sin{\left(6 x \right)}}{192} + \frac{{\color{red}{\left(6 \int{\cos{\left(4 x \right)} d x}\right)}}}{32}$$

Let $u=4 x$.

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

Thus,

$$\frac{5 x}{16} - \frac{15 \sin{\left(2 x \right)}}{64} - \frac{\sin{\left(6 x \right)}}{192} + \frac{3 {\color{red}{\int{\cos{\left(4 x \right)} d x}}}}{16} = \frac{5 x}{16} - \frac{15 \sin{\left(2 x \right)}}{64} - \frac{\sin{\left(6 x \right)}}{192} + \frac{3 {\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{16}$$

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

$$\frac{5 x}{16} - \frac{15 \sin{\left(2 x \right)}}{64} - \frac{\sin{\left(6 x \right)}}{192} + \frac{3 {\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{16} = \frac{5 x}{16} - \frac{15 \sin{\left(2 x \right)}}{64} - \frac{\sin{\left(6 x \right)}}{192} + \frac{3 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}}{16}$$

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

$$\frac{5 x}{16} - \frac{15 \sin{\left(2 x \right)}}{64} - \frac{\sin{\left(6 x \right)}}{192} + \frac{3 {\color{red}{\int{\cos{\left(u \right)} d u}}}}{64} = \frac{5 x}{16} - \frac{15 \sin{\left(2 x \right)}}{64} - \frac{\sin{\left(6 x \right)}}{192} + \frac{3 {\color{red}{\sin{\left(u \right)}}}}{64}$$

Recall that $u=4 x$:

$$\frac{5 x}{16} - \frac{15 \sin{\left(2 x \right)}}{64} - \frac{\sin{\left(6 x \right)}}{192} + \frac{3 \sin{\left({\color{red}{u}} \right)}}{64} = \frac{5 x}{16} - \frac{15 \sin{\left(2 x \right)}}{64} - \frac{\sin{\left(6 x \right)}}{192} + \frac{3 \sin{\left({\color{red}{\left(4 x\right)}} \right)}}{64}$$

Therefore,

$$\int{\sin^{6}{\left(x \right)} d x} = \frac{5 x}{16} - \frac{15 \sin{\left(2 x \right)}}{64} + \frac{3 \sin{\left(4 x \right)}}{64} - \frac{\sin{\left(6 x \right)}}{192}$$

Simplify:

$$\int{\sin^{6}{\left(x \right)} d x} = \frac{60 x - 45 \sin{\left(2 x \right)} + 9 \sin{\left(4 x \right)} - \sin{\left(6 x \right)}}{192}$$

Add the constant of integration:

$$\int{\sin^{6}{\left(x \right)} d x} = \frac{60 x - 45 \sin{\left(2 x \right)} + 9 \sin{\left(4 x \right)} - \sin{\left(6 x \right)}}{192}+C$$

$\int \sin^{6}{\left(x \right)}\, dx = \frac{60 x - 45 \sin{\left(2 x \right)} + 9 \sin{\left(4 x \right)} - \sin{\left(6 x \right)}}{192} + C$A