Integral of $$$8 \sin^{2}{\left(t \right)}$$$

Find $$$\int 8 \sin^{2}{\left(t \right)}\, dt$$$.

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=8$$$ and $$$f{\left(t \right)} = \sin^{2}{\left(t \right)}$$$:

$${\color{red}{\int{8 \sin^{2}{\left(t \right)} d t}}} = {\color{red}{\left(8 \int{\sin^{2}{\left(t \right)} d t}\right)}}$$

Apply the power reducing formula $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$ with $$$\alpha=t$$$:

$$8 {\color{red}{\int{\sin^{2}{\left(t \right)} d t}}} = 8 {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 t \right)}}{2}\right)d t}}}$$

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

$$8 {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 t \right)}}{2}\right)d t}}} = 8 {\color{red}{\left(\frac{\int{\left(1 - \cos{\left(2 t \right)}\right)d t}}{2}\right)}}$$

Integrate term by term:

$$4 {\color{red}{\int{\left(1 - \cos{\left(2 t \right)}\right)d t}}} = 4 {\color{red}{\left(\int{1 d t} - \int{\cos{\left(2 t \right)} d t}\right)}}$$

Apply the constant rule $$$\int c\, dt = c t$$$ with $$$c=1$$$:

$$- 4 \int{\cos{\left(2 t \right)} d t} + 4 {\color{red}{\int{1 d t}}} = - 4 \int{\cos{\left(2 t \right)} d t} + 4 {\color{red}{t}}$$

Let $$$u=2 t$$$.

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

Therefore,

$$4 t - 4 {\color{red}{\int{\cos{\left(2 t \right)} d t}}} = 4 t - 4 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}$$

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

$$4 t - 4 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}} = 4 t - 4 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}$$

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

$$4 t - 2 {\color{red}{\int{\cos{\left(u \right)} d u}}} = 4 t - 2 {\color{red}{\sin{\left(u \right)}}}$$

Recall that $$$u=2 t$$$:

$$4 t - 2 \sin{\left({\color{red}{u}} \right)} = 4 t - 2 \sin{\left({\color{red}{\left(2 t\right)}} \right)}$$

Therefore,

$$\int{8 \sin^{2}{\left(t \right)} d t} = 4 t - 2 \sin{\left(2 t \right)}$$

Add the constant of integration:

$$\int{8 \sin^{2}{\left(t \right)} d t} = 4 t - 2 \sin{\left(2 t \right)}+C$$

$$$\int 8 \sin^{2}{\left(t \right)}\, dt = \left(4 t - 2 \sin{\left(2 t \right)}\right) + C$$$A