Integral of $$$\sin{\left(\frac{3 \pi t}{2} \right)}$$$

Find $$$\int \sin{\left(\frac{3 \pi t}{2} \right)}\, dt$$$.

Let $$$u=\frac{3 \pi t}{2}$$$.

Then $$$du=\left(\frac{3 \pi t}{2}\right)^{\prime }dt = \frac{3 \pi}{2} dt$$$ (steps can be seen »), and we have that $$$dt = \frac{2 du}{3 \pi}$$$.

Therefore,

$${\color{red}{\int{\sin{\left(\frac{3 \pi t}{2} \right)} d t}}} = {\color{red}{\int{\frac{2 \sin{\left(u \right)}}{3 \pi} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{2}{3 \pi}$$$ and $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$${\color{red}{\int{\frac{2 \sin{\left(u \right)}}{3 \pi} d u}}} = {\color{red}{\left(\frac{2 \int{\sin{\left(u \right)} d u}}{3 \pi}\right)}}$$

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

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

Recall that $$$u=\frac{3 \pi t}{2}$$$:

$$- \frac{2 \cos{\left({\color{red}{u}} \right)}}{3 \pi} = - \frac{2 \cos{\left({\color{red}{\left(\frac{3 \pi t}{2}\right)}} \right)}}{3 \pi}$$

Therefore,

$$\int{\sin{\left(\frac{3 \pi t}{2} \right)} d t} = - \frac{2 \cos{\left(\frac{3 \pi t}{2} \right)}}{3 \pi}$$

Add the constant of integration:

$$\int{\sin{\left(\frac{3 \pi t}{2} \right)} d t} = - \frac{2 \cos{\left(\frac{3 \pi t}{2} \right)}}{3 \pi}+C$$

$$$\int \sin{\left(\frac{3 \pi t}{2} \right)}\, dt = - \frac{2 \cos{\left(\frac{3 \pi t}{2} \right)}}{3 \pi} + C$$$A