Integral of $$$\sin^{4}{\left(\theta \right)}$$$

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

Apply 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 $$$\alpha=\theta$$$:

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

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

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, d\theta = c \theta$$$ with $$$c=3$$$:

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

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

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

Let $$$u=2 \theta$$$.

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

Therefore,

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

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

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

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

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

Recall that $$$u=2 \theta$$$:

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

Let $$$u=4 \theta$$$.

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

Thus,

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

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

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

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

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

Recall that $$$u=4 \theta$$$:

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

Therefore,

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

Add the constant of integration:

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

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